please help me I need help ​

Answer:

1. Degree = 1, monomial

2. Degree = 2, monomial

3. degree = 2, trinomial

4. Degree = 2, binomial

5. Degree  = 2, binomial

Step-by-step explanation:

Answer:

375 guests

Step-by-step explanation:

costs: 1500 + 8g

income: 12g

They must be equal

1500+8g = 12g

Subtract 8g from each side

1500 +8g-8g =12g-8g

1500= 4g

Divide by 4

1500/4 = 4g/4

375 = g

Answer:

D. isosceles triangle

Step-by-step explanation:

Ed22

A triangle with at least 1-fold reflectional symmetry is isosceles triangle, option D is correct.

A triangle is a three-sided polygon that consists of three edges and three vertices.

An isosceles triangle will always have at least 1-fold reflectional symmetry.

This is because an isosceles triangle has two congruent sides and two congruent angles.

If we draw a perpendicular bisector of the base (the side that is not congruent), it will bisect the base and the angle opposite to the base.

This means that the triangle is symmetric with respect to this line of reflection, which is the line of symmetry.

Therefore, the correct answer is isosceles triangle.

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Answer:

C.

84

Step-by-step explanation:

This question is solved using proportions.

From the sample:

11 + 3 = 14 out of 13 + 11 + 3 = 27 would rate the test something other than easy.

Out of 162:

Applying the rule of three:

14 - 27

x - 162

Applying cross multiplication:

[tex]27x = 14*162[/tex]

[tex]x = \frac{14*162}{27}[/tex]

[tex]x = 84[/tex]

Thus the correct answer is given by option C.

Answer:

I hope this helps

Step-by-step explanation:

Answer:

s snsnnssjsjjsnsnsjs

es 17 es

Area of parallelogram = b × h

Base = x + 7

Height = x + 3

ATQ

Area = (x + 7) ( x + 3)

x² + 3x + 7x + 21

x² + 10x + 21

Answered by Gauthmath must click thanks and mark brainliest

Answer:

distance=√[(x2-x1)²+(y2-y1)²]

√[{6-(-2)}²+ (-8-(-4))²]

√(64+16)

√[100]

10

Points given

Distance:-

[tex]\\ \sf \longmapsto \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

[tex]\\ \sf \longmapsto \sqrt{(-8+4)^2+(6+2)^2}[/tex]

[tex]\\ \sf \longmapsto \sqrt{(-4)^2+(8)^2}[/tex]

[tex]\\ \sf \longmapsto \sqrt{64+16}[/tex]

[tex]\\ \sf \longmapsto \sqrt{80}[/tex]

[tex]\\ \sf \longmapsto 8.4[/tex]

9514 1404 393

Answer:

  KM = 8

Step-by-step explanation:

The ratio of long side to short side is the same for all of the triangles in this geometry:

  KM/ML = JM/KM

  KM² = ML·JM = 4(20-4) = 64

  KM = √64 = 8

The length of KM is 8 units.

Answer:

a. 24.12 ft³/hr b. 0.0768 ft/hr

Step-by-step explanation:

a. Find the rate at which the volume of water in the pool is increasing at time t=6 hours.

The net rate of change of volume of the cylinder dV/dt = volume flow rate in - volume flow rate out

Since volume flow rate in = P(t) and volume flow rate out = R(t),

dV/dt = P(t) - R(t)

[tex]\frac{dV}{dt} = P(t) - 18e^{0.04t}[/tex]

We need to find the rate of change of volume when t = 6.

From the table when t = 6, P(6) = 47 ft³/hr

Also, substituting t = 6 into R(t), we have R(6)

[tex]\frac{dV}{dt} = P(t) - 18e^{0.04t}[/tex]

[tex]\frac{dV}{dt} = 47 - 18e^{0.04X6}\\\frac{dV}{dt} = 47 - 18e^{0.24}\\\frac{dV}{dt} = 47 - 18 X 1.27125\\\frac{dV}{dt} = 47 - 22.882\\\frac{dV}{dt} = 24.118 ft^{3}/hr[/tex]

dV/dt ≅ 24.12 ft³/hr

So, the rate at which the water level in the pool is increasing at t = 6 hours is 24.12 ft³/hr

b. How fast is the water level in the pool rising at t=6 hours?

Since the a rate at which the water level is rising is dV/dt and the volume of the cylinder is V = πr²h where r = radius of cylinder = 10 ft and h = height of cylinder = 5 feet

dV/dt = d(πr²h)/dt = πr²dh/dt since the radius is constant and dh/dt is the rate at which the water level is rising.

So, dV/dt = πr²dh/dt

dh/dt = dV/dt ÷ πr²

Since dV/dt = 24.12 ft³/hr and r = 10 ft,

Substituting the values of the variables into the equation, we have that

dh/dt = dV/dt ÷ πr²

dh/dt = 24.12 ft³/hr ÷ π(10 ft)²

dh/dt = 24.12 ft³/hr ÷ 100π ft²

dh/dt = 0.2412 ft³/hr ÷ π ft²

dh/dt = 0.2412 ft³/hr

dh/dt = 0.0768 ft/hr

So, the arate at which the water level is rising at t = 6 hours is 0.0768 ft/hr

There are 50 different possible debating teams that could be selected as obtained using COMBINATION.

Since there are EQUAL number of juniors and seniors ;

Then we have 5 of each.

Here, the order of arrangement DOES NOT matter, Hence, we use COMBINATION

since the team MUST contain 4 SENIORS and 2 JUNIORS

4 Seniors from 5  = 5C4 = 5

2 Juniors from 5 = 5C2 = 10

Hence, (5C4 * 5C2) = 5 * 10 = 50

Hence, there are 50 different possible debating teams that could be selected.

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Answer:

Step-by-step explanation:

3\4 bc on a nuberline it would be 3 3\4

                                                        I--I----(etc)

so yeah hope i helped

Answer:

-1/2

Step-by-step explanation:

We can use the slope formula

m = ( y2-y1)/(x2-x1)

   = ( -2 - -4)/( -11 - -7)

   =( -2+4)/( -11+7)

   = 2 / -4

  = -1/2

ANSWER: The glass window will cost $252.00 for window replacement.

EXPLANATION:

The area of a square is given by the formula:

s^2

where s is the side length.

If we were to replace the glass window, and it will cost $7.00 per square foot, the total price will be:

FIND AREA OF THE GLASS WINDOW:

6^2 = 36 square feet

$7 per square foot so:

36 x 7 = 252

Therefore it will cost $252.00 to replace the glass window.

9514 1404 393

Answer:

  E.  384 cu ft

Step-by-step explanation:

The sum of length and width is half the perimeter, so the width is ...

  (40/2) -12 = 8 . . . feet

The depth is half that, so is 8/2 = 4 feet.

The volume is ...

  V = LWH = (12 ft)(8 ft)(4 ft) = 384 ft³

9514 1404 393

Answer:

  10 and 24

Step-by-step explanation:

We know that some of the Pythagorean triples that appear in math problems are (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17).

These have (perimeter, area) values of (12, 6), (30, 30), (56, 84), (40, 60).

For some scale factor n, we want (p·n, a·n²) = (60, 120). Of the triangles listed, we see that the (5, 12, 13) triangle scaled by n=2 will satisfy the problem requirements. (30·2, 30·2²) = (60, 120)

The side lengths are 10 and 24.

__

Check

For the side lengths we found, the perimeter is 10+24+26 = 60; the area is 1/2(10)(24) = 120. The hypotenuse is 2×13 = 26 = 24+2.

__

In the attached, one side is x, the other is y. The hypotenuse is (x+2). The square root equation comes from ...

  x² +y² = (x+2)²   ⇒   y² = (x² +4x +4) -x²   ⇒   y² = 4x +4 = 4(x +1)

_____

Additional comment

The graph shows the solution of the various constraints. At least, the combination of constraints will give a quadratic equation in x. They can be combined in a way that gives a cubic equation in x. Either way, we prefer the graphical or "guess and check" approach (above) as being easier to do.

Using the third equation in the attachment to write an expression for y, we have ...

  y = 58 -2x

Substituting that into the second equation gives ...

  (x(58 -2x)/2 = 120

  29x -x² = 120

  x² -29x +120 = 0

  (x -5)(x -24) = 0 . . . . x = 5 or 24.

The root x=5 is a legitimate solution to the pair of equations we chose to solve. The line y=58-2x intersects the hyperbola xy/2 = 120 in two places. However, (x, y) = (5, 48) does not satisfy the hypotenuse requirement that x+2 > y.

Answer:

Sorry I dont really understand wish I could help:(

Step-by-step explanation:

Answer:

[tex]\sqrt{10}[/tex]

[tex]\sqrt{5 } ^{2} + \sqrt{5 } ^{2} = x^{2}[/tex]

[tex]x^{2} =10[/tex]

Step-by-step explanation:

Answer:

64

Step-by-step explanation:

The Ratio of Blue to White drops is 2:8

16*4=64,... 16:64 Have a nice day!

Answer:

too complex:<

Step-by-step explanation:

120cm+120cm=240cm (2 squirrels + 2 air spaces)

90cm+90cm=180cm (2 rats + 2 air spaces)

240cm-180cm=(2 squirrels + 2 air spaces) - (2 rats + 2 air spaces)

                       =2 squirrels + 2 air spaces - 2 rats - 2 air spaces

                       =2 squirrels - 2 rats

                       =60cm

1 squirrel - 1 rat = 60cm divided by 2

                         = 30cm

120cm + 90cm = squirrel + air space + rat + air space

                        = 210cm

I've no idea!! This qn is too challenging!!

But i hope the above workings might help you in a way or another:>

The table is 105cm tall.

Two or more expressions with an Equal sign is called as Equation.

Let the table is x

Squirrel is y

Rat is z

From 1st diagram

x+y-z=120...(1)

From 2nd diagram

x+z-y=90...(2)

Add 1 and 2

x+y-z+x+z-y=120+90

2x=210

Divide both sides by 2

x=105

Hence, the table is 105cm tall.

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Answer:

The 90% confidence interval for the difference of proportions is (0.01775,0.18225).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

p1 -> 1993

20 out of 100, so:

[tex]p_1 = \frac{20}{100} = 0.2[/tex]

[tex]s_1 = \sqrt{\frac{0.2*0.8}{100}} = 0.04[/tex]

p2 -> 1997

10 out of 100, so:

[tex]p_2 = \frac{10}{100} = 0.1[/tex]

[tex]s_2 = \sqrt{\frac{0.1*0.9}{100}} = 0.03[/tex]

Distribution of p1 – p2:

[tex]p = p_1 - p_2 = 0.2 - 0.1 = 0.1[/tex]

[tex]s = \sqrt{s_1^2+s_2^2} = \sqrt{0.04^2 + 0.03^2} = 0.05[/tex]

Confidence interval:

[tex]p \pm zs[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

90% confidence level

So [tex]\alpha = 0.1[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].  

The lower bound of the interval is:

[tex]p - zs = 0.1 - 1.645*0.05 = 0.01775 [/tex]

The upper bound of the interval is:

[tex]p + zs = 0.1 + 1.645*0.05 = 0.18225 [/tex]

The 90% confidence interval for the difference of proportions is (0.01775,0.18225).

We know that,

[tex]EF=\dfrac{AD+BC}{2}[/tex]

which is

[tex]x=\dfrac{x-5+2x-4}{2}[/tex]

Now solve for x,

[tex]x=\dfrac{3x-9}{2}[/tex]

[tex]2x=3x-9[/tex]

[tex]x=9[/tex]

Since x is 9, the lengths are,

[tex]AD=x-5=9-5=\boxed{4}[/tex]

[tex]EF=x=\boxed{9}[/tex]

[tex]BC=2x-4=18-4=\boxed{14}[/tex]

Hope this helps :)

Answer:

no solution

Step-by-step explanation:

a negative number cannot be square rooted

Answer:

"not possible". no such thing as a negative squared number

Answer:

please ask in English

Step-by-step explanation:

then I can help

Answer:

Tom, 10.25

Step-by-step explanation:

Distance covered by Tom=4.5*(4 1/2+5)=81/4=42.75km

Distance covered by Jerry=6.5*(5)=32.5km

Tom is ahead from Jerry by 10.25km

Let the ratio be m:n

We know

[tex]\boxed{\sf (x,y)=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)}[/tex]

[tex]\\ \sf\longmapsto (4,5)=\left(\dfrac{7m+2n}{m+n},\dfrac{8m+3n}{m+n}\right)[/tex]

Now

.[tex]\\ \sf\longmapsto \dfrac{7m+2n}{m+n}=4\dots(1)[/tex]

[tex]\\ \sf\longmapsto \dfrac{8m+3n}{m+n}=5\dots(2)[/tex]

Adding both

[tex]\\ \sf\longmapsto \dfrac{7m+2n+8m+3n}{m+n}=4+5[/tex]

[tex]\\ \sf\longmapsto \dfrac{7m+8m+2n+3n}{m+n}=9[/tex]

[tex]\\ \sf\longmapsto \dfrac{15m+5n}{m+n}=9[/tex]

[tex]\\ \sf\longmapsto 15m+5n=9(m+n)[/tex]

[tex]\\ \sf\longmapsto 15m+5n=9m+9n[/tex]

[tex]\\ \sf\longmapsto 15m-9m=9n-5n[/tex]

[tex]\\ \sf\longmapsto 6m=4n[/tex]

[tex]\\ \sf\longmapsto \dfrac{m}{n}=\dfrac{6}{4}[/tex]

[tex]\\ \sf\longmapsto \dfrac{m}{n}=\dfrac{3}{2}[/tex]

[tex]\\ \sf\longmapsto m:n=3:2[/tex]

Option B is correct

Answer:

In mathematics, a real number is a value of a continuous quantity that can represent a distance along a line (or alternatively, a quantity that can be represented as an infinite decimal expansion).

Answer:

i will say methods try yourself:

How to Determine Maximum Value

How to Determine Maximum ValueIf your equation is in the form ax2 + bx + c, you can find the maximum by using the equation:

How to Determine Maximum ValueIf your equation is in the form ax2 + bx + c, you can find the maximum by using the equation:max = c - (b2 / 4a).

How to Determine Maximum ValueIf your equation is in the form ax2 + bx + c, you can find the maximum by using the equation:max = c - (b2 / 4a).The first step is to determine whether your equation gives a maximum or minimum. ...

How to Determine Maximum ValueIf your equation is in the form ax2 + bx + c, you can find the maximum by using the equation:max = c - (b2 / 4a).The first step is to determine whether your equation gives a maximum or minimum. ...-x2 + 4x - 2.

How to Determine Maximum ValueIf your equation is in the form ax2 + bx + c, you can find the maximum by using the equation:max = c - (b2 / 4a).The first step is to determine whether your equation gives a maximum or minimum. ...-x2 + 4x - 2.Since the term with the x2 is negative, you know there will be a maximum point.

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plzz follow and make brainlist

Answer:

[tex]y=\left(6\right)^{x}\ -3[/tex]

Step-by-step explanation:

Answer:

A

Step-by-step explanation:

I guess that is the answer