What are the solutions of the equation 3x^2+6x-24=0

Answer:

[tex]163 -y =-5\\Collect\:Like\:terms\\163+5 = y\\Simplify\\168 =y\\\\y = 168[/tex]

Hello There!

[tex]163-y=-5[/tex]

To solve your equation, you can just change the -5 to 5 and move it to where y is. After that, change the minus sign to addition.

[tex]163+5=y[/tex]

Now all you have to do is sum it up.

[tex]163+5=168[/tex]

So y = to 168

So your answer is

[tex]163-168=-5[/tex]

Hope this Helps!

[tex]10[/tex] divisions between $15.59$ and $15.6$ so each division is $\frac{15.60-15.59}{10}=0.001$

A is 5 division from $15.59$, so, A is $15.59+5\times 0.001=15.595$

similarly, C is 4 division behind $15.59$ so it is $15.590-4\times0.001=15.586$

and B is $15.601$

Answer:

Answer is attached below :)

Answer:

Hey there!

Simple interest formula: I=PRT

I=700(8.25)(0.25)

I=1443.75

Hope this helps :)

Answer:

Step-by-step explanation:

I = PRT

I = 700(0.0825)(1/4) = 14.44

Because the interest is usually in percentage and it's impossible to have 825% as your interest rate. So the actual interest rate has to be 0.0825.

The formula above calculated the interest, if you want the total, you will need to add 700 to that number.  

Here's a small quick example of the formula that should help.

Answer:

Sample size n [tex]\simeq[/tex] 1269.15

Step-by-step explanation:

From the information given ,

At 99% of confidence interval,

the level of significance ∝ = 1 - 0.99

the level of significance ∝ =  0.01

the critical value for 99% of confidence interval is:

[tex]\mathtt{\dfrac{\alpha }{2} = \dfrac{0.01}{2}}[/tex]

= 0.005

[tex]\mathtt {z_{\alpha/2} = z_{0.005/2} }[/tex]

The value for z from the standard normal tables

= 2.58

The Margin of error E= 3% = 0.03

The formula to  determine the sample size n used can be expressed as follows:

[tex]\mathtt { n = (\dfrac{z_{\alpha/2}}{E})^2 \ \hat p (1 - \hat p) }[/tex]

where;

[tex]\mathtt{\hat p }[/tex] = 22% = 0.22

Then:

[tex]\mathtt { n = (\dfrac{2.58}{0.03})^2 \ \times 0.22 \times (1 - 0.22) }[/tex]

[tex]\mathtt { n = (86)^2 \ \times 0.22 \times (0.78) }[/tex]

[tex]\mathtt { n = 7396 \ \times 0.22 \times (0.78) }[/tex]

n = 1269.1536

Sample size n [tex]\simeq[/tex] 1269.15

Answer:

D. the bottom one is the answer, because hyperbola is two curves that curve infinitely

Answer:

Step-by-step explanation:

The first thing you want to do is to factor in any quadratic equation.

So, -10(x^2-25)

Now, we see this is a special case, whenever we see a equation in this case, x^2 - b^2, we factor it to this (x+b)(x-b)

So, -10(x+5)(x-5)

x = -5 and x = 5

Answer:

There is a typo near the equal sign.

There can be two different answers if we think that = sign as + or -.

First way: Making = as +

=> 10 - [14 + (3+4) x 2] +3

=> 10 - [14 + 7 x 2] + 3

=> 10 - [14 + 14] + 3

=> 10 - 28 + 3

=> 10 + 3 - 28

=> 13 - 28

=> -15

=> So, -15 is the answer if we consider "=" sign as "+" sign.

Second way: Making = as -

=> 10 - [14 - (3+4) x 2] + 3

=> 10 - [14 - 7 x 2] + 3

=> 10 - [14 - 14] + 3

=> 10 - 0 + 3

=> 10 + 3

=> 13

=> So, 13 is the answer if we consider "=" sign as "-" sign.

Answer:

5:8

Step-by-step explanation:

If I understand your question correctly, we have 1500/2400=15/24=5/8, so we have Lynn:Bo is 5:8, however, in the future please be more clear.

What is "estimates roof jon"? And, instead of saying "ratio to lynn to bo" say "What is the ratio of the estimates?" or whatever you're asking. If this answer is wrong, you only have yourself to blame.

Answer:

The  probability is  [tex]P(X \le 0.305 ) = 0.01795[/tex]

Step-by-step explanation:

From the question we are told that

      The population mean is  [tex]\mu = 0.966 \ grams[/tex]

       The standard deviation is  [tex]\sigma = 0.315 \ grams[/tex]

Given that the amounts of nicotine in a certain brand of cigarette are normally distributed

    Then the probability of randomly selecting a cigarette with 0.305 grams of nicotine or less is mathematically represented as

        [tex]P(X \le 0.305 ) = 1 - P(X > 0.305) = 1 - P(\frac{X - \mu }{\sigma } > \frac{0.305 - \mu }{\sigma } )[/tex]

Generally

              [tex]\frac{X - \mu }{\sigma } = Z (The \ standardized \ value \ of X )[/tex]

So  

      [tex]P(X \le 0.305 ) = 1 - P(X > 0.305) = 1 - P(Z > \frac{0.305 - 0.966 }{0.315} )[/tex]

      [tex]P(X \le 0.305 ) = 1 - P(X > 0.305) = 1 - P(Z >-2.0984 )[/tex]

From the z-table(reference calculator dot  net  ) value of   [tex]P(Z >-2.0984 ) =0.98205[/tex]

So  

     [tex]P(X \le 0.305 ) = 1 - P(X > 0.305) = 1 - 0.98205[/tex]

=>  [tex]P(X \le 0.305 ) = 1 - P(X > 0.305) = 0.01795[/tex]

=> [tex]P(X \le 0.305 ) = 0.01795[/tex]

Answer:

Find answer below

Step-by-step explanation:

f(x)=2x-6

Domain of 2x-6: {solution:-∞<x<∞, interval notation: -∞, ∞}

Range of 2x-6: {solution:-∞<f(x)<∞, interval notation: -∞, ∞}

Parity of 2x-6: Neither even nor odd

Axis interception points of 2x-6: x intercepts : (3, 0) y intercepts (0, -6)

inverse of 2x-6: x/2+6/2

slope of 2x-6: m=2

Plotting : y=2x-6

Answer:

D

Step-by-step explanation:

[tex]4(x^2+3)-2y\\\\=4((-6)^2+3)-2(\frac{-1}{2} )\\\\=4(36+3)+1\\\\=4(39)+1\\\\=156+1\\\\=157[/tex]

The value of the expression 4(x² + 3) - 2y is 157, when x = -6 and y = -1/2.

An algebraic expression is consists of variables, numbers with various mathematical operations,

The given expression is,

4(x² + 3) - 2y

Substitute x = -6 and y = -1/2 to find the value of expression,

= 4 ((-6)² + 3) - 2(-1/2)

= 4 (36 + 3) + 1

= 4 x 39 + 1

= 156 + 1

= 157

The required value of the expression is 157.

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Step-by-step explanation:

( secA + 1)( sec A - 1)

Using the expansion

( a + b)( a - b) = a² - b²

Expand the expression

We have

sec²A + secA - secA - 1

That's

sec² A - 1

From trigonometric identities

So we have the final answer as

As proven

Hope this helps you

Step-by-step explanation:

Here,

LHS

= (SecA+1)(secA -1)

[tex] = {sec}^{2} A - 1[/tex]

[tex]{as{a}^{2} - {b}^{2} =(a + b)(a - b)[/tex]

Now, we have formula that:

[tex] {sec}^{2} \alpha - {tan \alpha }^{2} = 1[/tex]

[tex] {tan}^{2} \alpha = {sec }^{2} \alpha - 1[/tex]

as we got ,

[tex] = {sec}^{2} A- 1[/tex]

This is equal to:

[tex] = {tan}^{2} A[/tex]

= RHS proved.

Hope it helps....

Answer:

1 square foot is the answer

Answer:

1 ft^2

Step-by-step explanation:

We know 12 inches = 1 ft

12 inches by 12 inches

1 ft by 1 ft

The area is 1 * 1 = 1 ft^2

Answer:

72[tex]\sqrt{3}[/tex] units²

Step-by-step explanation:

The area (A) of the triangle is calculated as

A = [tex]\frac{1}{2}[/tex] bh ( b is the base and h the perpendicular height )

Here b = ST = a = 12 and h = RS

To calculate RS use the tangent ratio in the right triangle and the exact value

tan60° = [tex]\sqrt{3}[/tex] , thus

tan60° = [tex]\frac{opposite}{adjacent}[/tex] = [tex]\frac{RS}{ST}[/tex] = [tex]\frac{RS}{12}[/tex] = [tex]\sqrt{3}[/tex] ( multiply both sides by 12 )

RS = 12[tex]\sqrt{3}[/tex]

Thus

A = [tex]\frac{1}{2}[/tex] × 12 × 12[tex]\sqrt{3}[/tex] = 6 × 12[tex]\sqrt{3}[/tex] = 72[tex]\sqrt{3}[/tex] units²

Answer:

(-7, -4)  which is your answer D in the list of options

Step-by-step explanation:

The center of the circle should be located half way in between the given points on the plane.

Then the center ahs to be located half way for the x coordinates of both points:

half way between -12 and -2 (notice that there is a difference of 10 units between them), therefore half way would be at 5 units to the right from the furthest point, that is  -12 + 5 = -7

Similarly, for the y coordinate, we see that the difference is between -5 and -3 (a difference of two units) therefore the center point will be located half way  (that is one unit) up from the lowest y coordinate: -5 + 1 = -4

Then the center of the circle is located at (-7, -4)

Answer:

12 units

Step-by-step explanation:

Since all of the sides of a rhombus are congruent, JK = JM which means:

2x + 4 = 3x

-x = -4

x = 4 so 3x = 3 * 4 = 12

Answer:

fucuvucybycych tcy bic ttx TV ubtx4 cub yceec inivtxr xxv kb

Step-by-step explanation:

t tcextvtcbu6gt CNN tx r.c tct yvrr TV unu9gvt e tch r,e xxv t u.un4crcuv3cinycycr xxv yctzrctvtcrzecycyvubr xiu nyfex tut uhyh

Answer:

Question 18: B. 104

Question 19: [tex] x = \frac{3}{2} [/tex]

Step-by-step Explanation:

Question 18:

Step 1: express the inverse relationship with an equation

[tex] y = \frac{k}{x^2} [/tex] ,

where k is constant

y = 26 when x = 4,

Constant, k, = [tex] y*x^2 = k [/tex]

[tex] k = 26*4^2 = 416 [/tex]

The equation would be [tex] y*x^2 = 416 [/tex]

Step 2: use the equation to find y when X = 2.

[tex] y*x^2 = 416 [/tex]

[tex] y*2^2 = 416 [/tex]

[tex] y*4 = 416 [/tex]

Divide both sides by 4

[tex] \frac{y*4}{4} = \frac{416}{4} [/tex]

[tex] y = 104 [/tex]

Question 19:

[tex] \frac{x}{3} = \frac{x + 2}{7} [/tex]

Cross multiply

[tex] x(7) = 3(x + 2) [/tex]

[tex] 7x = 3x + 6 [/tex]

Subtract 3x from both sides

[tex] 7x - 3x = 3x + 6 - 3x [/tex]

[tex] 4x = 6 [/tex]

Divide both sides by 4

[tex] \frac{4x}{4} = \frac{6}{4} [/tex]

[tex] x = \frac{3}{2} [/tex]

Answer: D.) 52

Explanation: I guessed and got it right lol

Answer:

34 grams

Step-by-step explanation:

If the remaining sample has 30.26 grams of radioactive substance, and 11% of it decayed, that means that 30.26 grams is 89% of the original. Let the original be x.

30.26=0.89x

Multiply both by one hundred

3026=89x

Divide both by 89

34=x

x=original, so the original was 34 grams.

Answer: $43,823.37

Step-by-step explanation:

Formula to calculate the accumulated amount earned on principal (P) at rate of interest (r) compounded daily after t years :

[tex]A=P(1+\dfrac{r}{365})^{365t}[/tex]

As per given , we have

P= $ 30,700

r= 8.9 % = 0.089

t= 4 years

[tex]A=30700(1+\dfrac{0.089}{365})^{365(4)}\\\\=30700(1+0.0002438)^{365(4)}\\\\=30700(1.0002438)^{1460}\\\\=30700(1.42747138525)\\\\=43823.3715272\approx43823.37[/tex]

Hence, the amount at the end of 4 years would be $43,823.37 .

Answer:

2 servings of salad and 1 serving of soup

Step-by-step explanation:

In the given scenario the aim is to minimise the fat content of the food combination.

Fat content of soup is 3mg while fat content of salad is 2 mg.

Using Soup as 0 and Salad as 2 will not give the required vitamin content

The logical step will be to keep servings of soup to the minimum.

Let's see some combinations of salad and soup. Keeping serving of soup to the minimum of 1

1. 1 serving of salad and one serving of soup will contain 10 mg of vitamin A, 7 mg of vitamin B complex, and 3 mg of fat.

This will not work because amount of vitamin B complex is not up to 10 mg

2. 2 servings of salad and 1 serving of soup. Will contain 14 mg of vitamin A, 12 mg of vitamin B, and 7 mg of fat

This is the best option as we have amount of vitamin A and vitamin B complex in adequate quantity.

Also fat is lowest in this combination because soup the food with highest fat content is at minimum amount of one serving

Answer:

36

step by step

given length=6

so area of square is given by s2 i.e 6^2

=6×6

=36 (Ans)

Answer: The AP = 1, ⁷/₂, 6, ¹⁷/₂, 11 ..............

Step-by-step explanation:

From the first statement,

S₅ = ⁵/₂(2a + ( n - 1 )d }    = 30

       5(2a  + 4d )d           = 60

       10a    + 20d             = 60

reduce to lowest term to easy calculation by dividing through by 10

                      a + 2d       = 6 -----------------------------------1

second statement

sum of the next 4 terms inclusive

T₉  = ⁹/₂(2a + 8d )   = 69

        9(2a  + 8d )    = 30 + 69

            18a + 72d   = 99 x 2

            18a + 72d   = 198

divide through by 18 to reduce to lowest time

                a  + 4d    = 11 ------------------------------------------2

Now solve the two equation simultaneously to find a and d

                 a + 2d = 6

                 a + 4d = 11

                     -2d  = -5

                          d = ⁵/₂.

Now substitute  for d to get a

               a  + 2(⁵/₂) = 6

                a + 5       = 6

                           a   = 6 - 5

                            a = 1.

Therefore the AP = 1 , ⁷/₂ , 6 , ¹⁷/₂ , 11 , ..............

The AP if, The sum of the first 5 terms of an AP is 30 and the sum of the four terms from T6 to T9 is 69, is 1, ⁷/₂, 6, ¹⁷/₂, 11, and so on.

An ordered collection of objects that allows repetitions is referred to as a sequence. It has members, just like a set does. The length of the sequence is determined by the number of items.

Given:

The sum of the first 5 terms of an AP is 30,

Write the equations as shown below,

S₅ = ⁵/₂(2a + ( n - 1 )d }  = 30

5(2a  + 4d )d = 60

10a    + 20d = 60

reduce to lowest term to easy calculation by dividing through by 10

a + 2d = 6

T₉  = ⁹/₂(2a + 8d )   = 69  (sum of the next 4 terms inclusive)

9(2a  + 8d )  = 30 + 69

18a + 72d  = 99 x 2

18a + 72d = 198

a  + 4d    = 11

Solve the equation as shown below,

d = ⁵/₂, and   a = 1.

Therefore, the AP = 1, ⁷/₂, 6, ¹⁷/₂, 11, and so on.

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Answer:

3%

Step-by-step explanation:

1. Set up the equation

6(0.18) + 12x = 18(0.08)

2. Simplify

1.08 + 12x = 1.44

3. Solve

12x = 0.36

x = 0.03

0.03 = 3%

Answer:

d=55(t+0.5)

Step-by-step explanation:

d=55(t+0.5)

Answer:

27.5

Step-by-step explanation:

Answer:

x = -70

Step-by-step explanation:

x/10 = -7

Multiply each side by 10

x/10*10 = -7*10

x = -70

Answer:

The minimum sample size is [tex]n = 2123[/tex]

Step-by-step explanation:

From the question we are told that

    The margin of error is  [tex]E = 0.028[/tex]

Given that the confidence level is  99% then the level of significance is evaluated as

        [tex]\alpha = 100 - 99[/tex]

        [tex]\alpha = 1 \%[/tex]

        [tex]\alpha =0.01[/tex]

Next we obtain the critical value of  [tex]\frac{ \alpha }{2}[/tex] from the normal distribution table

   The  value is  [tex]Z_{\frac{ \alpha }{2} } = 2.58[/tex]

Now let  assume that the sample proportion is  [tex]\r p = 0.5[/tex]

  hence  [tex]\r q = 1 - \r p[/tex]

=>            [tex]\r q = 0.50[/tex]

   Generally the sample size is mathematically represented as

                [tex]n =[ \frac{Z_{\frac{ \alpha }{2} }}{ E} ]^2 * \r p * \r q[/tex]

                [tex]n =[ \frac{2.58}{ 0.028} ]^2 * 0.5 * 0.5[/tex]

              [tex]n = 2123[/tex]

Hello, please consider the following.

We know that

[tex]a = p^{\frac{1}{3}}-p^{-\frac{1}{3}}\\\\=p^{\frac{1}{3}}-\dfrac{1}{p^{\frac{1}{3}}}[/tex]

And we can write that.

[tex](p-\dfrac{1}{p})^3=(p-\dfrac{1}{p})(p^2-2+\dfrac{1}{p^2})\\\\=p^3-2p+\dfrac{1}{p}-p+\dfrac{2}{p}-\dfrac{1}{p^3}\\\\=p^3-\dfrac{1}{p^3}-3(p-\dfrac{1}{p})[/tex]

It means that, by replacing p by [tex]p^{1/3}[/tex]

[tex](p^{1/3}-\dfrac{1}{p^{1/3}})^3=p-\dfrac{1}{p}-3(p^{1/3}-\dfrac{1}{p^{1/3}})\\\\\\\text{ So }\\\\a^3=p-\dfrac{1}{p}-3a\\\\<=>\boxed{ a^3+3a=p-\dfrac{1}{p} }[/tex]

Hope this helps.

Do not hesitate if you need further explanation.

Thank you