Find the distance between the lines with equations 7x-2y-2=0 and y=7/2x-10?

a. 2.47 b. 2.48 c. 2.5 d. 2.49

Answer:

i think

third option is correct.

cosY = 8 / 17, cotY = 8 / 15, secY = 17 / 8 is the required trigonometric functions of the same angle. Option C is correct

These are the equation that contains trigonometric operators such as sin, cos.. etc.. In algebraic operations.

Here,
cosY = 8 / 17, cotY = 8 / 15, secY = 17 / 8
Is the trigonometric function of the same angle,
cosY = 1 / SecY
8 / 17 = 1 / secY
secY = 17 / 8
Now,
P = √[17² - 8²]
P = 15
CotY = 8 / 15

Thus, trigonometric functions in option C are of the same angle.

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Answer:

29/14

Step-by-step explanation:

Start by working out the LCM of these (or any common denominator). A common denominator for 4 and 7 is 28 cuz 4 x 7 = 28.

3/4 = 21/28

4/7 = 16/28

3/4 = 21/28

(21+16+21)/28 = 58/28 which can simplify to 29/14

hi

Golden rule : you can only add or substract fraction if and only if they have the same denominator.

so two solutions :

you see that one denominator is a multiple of the second. so you multiply all the fraction by this multiple.

ex : 1/3 + 5/6

here I can easely convert 1/3 into a fraction with 6 as 6 is 3x2.

so : 1x2 = 2 and 3x2 = 6

1/3 is 2/6

so. 1/3 +5/6 = 2/6 +5/6

fraction have same denominator, I add numerators : 2/6 +5/6 =7/6

if it not that evident, you multiply fraction A up and down by denominator of fraction B.

and you multiply up and down fraction B by denomonator of fraction A

let' s see with your exemle :

3/4 +4/7+3/4 = ?

first I add fraction with same denominator :

3/4+3/4+4/7 = 6/4 +4/7

fraction A is 6/4.

I multiply up and down by denominator fractiok B which is " 7"

so. 6/4 = 6x7/4×7 = 42/28

let's apply method to fraction B whixh is 4/7 . let' multiply up and down by denominator of fraction A which is " 4"

so : 4/7 = 4x4 /7x4 = 16/28

now we have :

6/4 +4/7 = 42/28 +16/28

both my fractions have same denominator "28" so I add numerator :

42/28 +16/28 = 58/28

Now must wonder if 58/28 can be simplify. Good trick is to decompose numbers, and in prime number is the best way to do

58 = 2x29

28 = 7x2x2

so 58/28= (2×29)/ (7x2x2)

when there is same number up and down I can cross it out .

so here one "2" up and down can be rule out

so 58/28 = 2x29 /7×2×2 = 29/7x2 =29/14

I can not symplify more, so calculus is over.

in short :

3/4+4/7+3/4 = 6/4+4/7

= 42/28 +16/28

=58/28

= 29/14

Answer:

(-4, -1)

Step-by-step explanation:

The solution to the system of equations on a graph is the point in which they intersect. Therefore, the solution is (-4, -1).

Answer:

(3x + 1)(x + 6)

3x^2 + 19x + 6

3x^2 + 18x + x + 6

3x^2 + 18x + x + 6

3x(x + 6) + x + 6

3x(x + 6) + x + 6

(3x + 1)(x + 6)

Haven't done this in a pretty long time so I might be wrong

Answer:

The 6th term is 3072.

Step-by-step explanation:

[tex]\to \bold{|x-5.67|=0.02}\\\\[/tex]

heaviest[tex]\bold{=5.69\ g\\\\}[/tex]

lightest[tex]\bold{=5.65\ g\\\\}[/tex]

Evaluating each expression:  

[tex]\to \bold{q = -8}\\\\ \to \bold{r = -6}\\\\ \to \bold{t=3}[/tex]

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Answer:

Step-by-step explanation:

3) Surface area of cube = 6a² where a is side of the cube.

a)  a = 6 cm

Surface area of cube = 6 * 6*6 = 216 cm²

b) a = 4.5 cm

Surface area of cube = 6 * 4.5 * 4.5 = 121.5 cm²

4)  Area of cuboid = A = 2(lb + bh + hl)

a) l = 10 cm ; h = 6 cm  ; A = 376 cm²

  A = 376

2(lb + bh + hl) = 376

2(10b + 6b + 60) = 376

2*( 16b + 60) = 376

 2*16b + 2*60 = 376

     32b + 120 = 376

        32b  = 376 - 120

          32b = 256

             b = 256 / 32

b = 8 cm

5) Surface area  of cube = 150 cm²

6a² = 150

Divide both sides by 6

a² = 150/6

a² = 25 = 5*5

a = 5 cm

6)  l = 15 cm ; b = 12 cm ; h = 10cm

Lidless. so area of top portion is not included

Surface area of lidless cuboid = 2(hb + hl) + lb

                                                  = 2*(10*12 + 10*15) + (15*12)

                                                  = 2*(120 + 150) + 180

                                                  = 2* 270 + 180

                                                  = 540 + 180

                                                   = 720 cm²

Step-by-step explanation:

Q5:Find the length of each side of the cube whose Total surface area are given below:

(B)2400cm^2

Solution:

T.S.A= 6a^2 [ we have to find a]

2400=6a^2

2400/6=a^2

400=a^2

a=squareroot of 400

=20

So, 20cm is the length of a cube.

Answer:

4x + 1

Step-by-step explanation:

(8x + 2) - (4x + 1)

subtract the terms that are common,,

8x-4x=4x

2-1=1

therefore, it would be 4x + 1

if you need more help with these, feel free to sub to gauthmath on subreddit

Answer:

A, it's the only one with a number greater then 1

Answer: $1.1x

Step-by-step explanation:

At the end of year 1, it will increase by 10%:

[tex]x+0.1x=x(1+0.1)=1.1x[/tex]

Answer:

y= - 1.4

x= 0.2

Step-by-step explanation:

2x-Y= -1

x-2y=4

Answer:

The conic section for the equation 5x^2 - y = 12 is parabola

Answer:

a.2×1= 2 + 3×1=3 +4×2=8 now we have to put this ans and the ans is 13.

b. 5×1×2=10 + 3×1×2=6+ 7×2×2=28 now also we have to put this ans and this ans is 13

The depth of the mirror of the cross-section of a searchlight would be 8.33 cm if the light bub has a vertex at 3 cm and the mirror is 20 centimeters across at the origin.

A cross-section is perpendicular to the axis of the symmetry goes through the vertex of the parabola. The cross-sectional shape of the mirrored section of most searchlights or spotlights is parabolic.

Given:

a = 3

y = [tex]\frac{20}{2}[/tex] cm = 10 cm

Solution:

from the equation [tex]y^{2} = 4ax[/tex]

[tex]y^{2} = 4*3*x\\ y^{2} = 12x[/tex]

putting x, 10 cm in the equation

[tex]x=\frac{10^{2} }{12} \\\\x= \frac{100}{12} \\\\x= 8.33 cm[/tex]

thus, the depth of the mirror would be - 8.33 cm

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Answer:

According to me,

Answer:

hello,

answer C

Step-by-step explanation:

area of the rectangle=2b*b=2b²

area of the half cercle= [tex]\\\dfrac{\pi*(\frac{c}{2})^2}{2} =\dfrac{\pi*c^2}{8} \\[/tex]

Area in blue= 2b²-πc²/8

Answer:

37.5%

Step-by-step explanation:

150/450 = .375 = 37.5%

The percentage is 150 grams out of 400 grams will be 37.5%.

The quantity of anything is stated as though it were a fraction of a hundred. A quarter of 100 can be used to express the ratio. Per 100 is what the term percent signifies. The symbol '%' is used to symbolize it.

The percentage is given as,

Percentage (P) = [Initial value - Final value] / Initial value x 100

It is given that the difference between the initial and final value is 15 grams and the initial value is 400 grams.

Then the percentage is 150 grams out of 400 grams will be

P = (150 / 400) x 100

P = 0.375 x 100

P = 37.5%

Thus, the percentage is 150 grams out of 400 grams will be 37.5%.

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Answer:

5.53°

Explanation:

Law of sines.

Considering only this customer's premiums, deductible, and claims, the break-even period after which the customer could make a $10,000 claim for property damage without the insurance company suffering a loss is 15 months.

At this break-even period, the company will not make a profit or a loss.

Data and Calculations:

Monthly insurance premium payable = $600 (assumed to be paid monthly)

Deductible on the insurance policy = $1,000

Minimum coverage = 30/50/10

Coverage for injury or death to one person = $30,000

Coverage for injury or death to two or more people = $50,000

Coverage for damage to property = $10,000

Average accidents every year = 2

Total claims cost to the insurance before deductibles = $40,000 ($30,000 + $10,000)

Break-even period = 15 ($10,000 - $1,000)/$600

This break-even period is computed by deducting the deductible from the claim that the customer makes, and then dividing the result by the monthly insurance premium.

Thus, the customer could make a $10,000 claim for property damage without the insurance company suffering a loss after 15 months' break-even period.  

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Answer:

a + b =1

Step-by-step explanation:

Write the given quadratic as x^2 – 1x – 90 = 0.  This factors into

(x - 10)(x + 9) = 0, and so the roots/solutions are {-9, 10}.

If we call -9 "a" and 10 "b," then the sum a + b is -9 + 10, or a + b =1.

Answer: x=5, y=-1

Hope this helps

Step-by-step explanation:

0 10

0-6

this is the answer. I am sure

The value of numbers are 6 and 10 which are a distance of 2 units from 8 on a number line.

Number line is a horizontal line where numbers are marked at equal intervals one after another form smaller to greater.

We have to given that;

Number line is shown in figure.

And, We have to find the value of numbers which are a distance of 2 units from 8 on a number line.

Now,

We can find the value of numbers as;

⇒ | 8 - 2 | = 6

⇒ | 8 + 2 | = 10

Therefore, The value of numbers are 6 and 10 which are a distance of 2 units from 8 on a number line.

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Answer:

x-intercepts = (x ≈ -0.87), (x ≈ 6.87)

Step-by-step explanation:

let x = 0, in equation for x-intercepts

solve for x using the

quadratic formula

x = −6 ± √ 36 + 24

               2x

= −6 ± √60

         −2x

 = −6 ± 2 √ 15

           −2           = 3 ± √15

9514 1404 393

Answer:

Step-by-step explanation:

The GCF can be no larger than the difference between the numbers. That would be the first value you want to check to see if it is a factor of the numbers.

The difference is 10, which is a factor of both 40 and 50.

  GCF(40, 50) = 10

__

You can figure the GCF pairwise, as GCF(14, 56) = 14; then GCF(14, 63) = 7. Or, you can look at the smallest difference between any pair of numbers. Here, that is 63 -56 = 7, which is a factor of all three numbers.

  GCF(14, 56, 63) = 7

_____

Additional comment

Euclid's algorithm has you compute the remainder from division of the largest by the smallest. When that remainder is non-zero, it replaces the largest, and you repeat. When the remainder is zero, the smallest is the GCF.

For example, let's look at the GCF of 14 and 63.

  63/14 = 4 r 7

  14/7 = 2 r 0 . . . . . 7 is the divisor, so is the greatest common factor

Answer:

Solution given:

Right angled triangle ABC is drawn where <C=[tex]\theta[/tex]

we know that

[tex]\displaystyle Sin\theta=\frac{opposite}{hypotenuse} =\frac{AB}{AC}[/tex]

[tex]\displaystyle Cos\theta=\frac{adjacent}{hypotenuse}=\frac{BC}{AC} [/tex]

Now

left hand side

[tex] \displaystyle {sin}^{2} \theta + {cos}^{2} \:\theta[/tex]

Substituting value

[tex](\frac{AB}{AC})²+(\frac{BC}{AC})²[/tex]

distributing power

[tex]\frac{AB²}{AC²}+\frac{BC²}{AC²}[/tex]

Taking L.C.M

[tex]\displaystyle \frac{AB²+BC²}{AC²}[/tex]....[I]

In ∆ABC By using Pythagoras law we get

[tex]\boxed{\green{\bold{Opposite²+adjacent²=hypotenuse²}}}[/tex]

AB²+BC²=AC²

Substituting value of AB²+BC² in equation [I]

we get

[tex]\displaystyle \frac{AC²}{AC²}[/tex]

=1

Right hand side

Step-by-step explanation:

I think you first have to calculate a°,this can be done by first finding c°

c°=180-125

=55°

opposite angles in a quadrilateral add up to 180

therefore a°=180-55

=125

therefore angle d

180-125

=55

I hope this helps

Answer:

D

Step-by-step explanation:

Answer:

It's D on there

Step-by-step explanation:

(a) Let x = 3.12333…. Then 100x = 312.333… and 1000x = 3123.333…, so that

1000x - 100x = 3123.333… - 312.333…

==>   900x = 2811

==>   x = 2811/900 = 937/300

(b) x = 4.38555…   ==>   100x = 438.555… and 1000x = 4385.555…

==>   900x = 3947

==>   x = 3947/900

(c) x = 37.222…   ==>   10x = 372.222…

==>   9x = 335

==>   x = 335/9

(d) x = 16.292929…   ==>   100x = 1629.292929…

==>   99x = 1613

==>   x = 1613/99

(e) x = 2.333…   ==>   10x = 23.333…

==>   9x = 21

==>   x = 21/9 = 7/3

Answer:

Step-by-step explanation:

12

Answer:

[tex]A' = (-2,2)[/tex]

[tex]B' = (2,2)[/tex]

[tex]C' = (2,-2)[/tex]

Step-by-step explanation:

Given

[tex]A = (-6,6)[/tex]

[tex]B = (6,6)[/tex]

[tex]C = (6,-6)[/tex]

[tex]k = \frac{1}{3}[/tex]

Required

The new coordinates

To do this, we simply multiply the coordinates of [tex]\triangle ABC[/tex] by the factor of dilation.

i.e.:

[tex]A' = A * k[/tex]

[tex]B' = B * k[/tex]

[tex]C' = C * k[/tex]

So, we have:

[tex]A' = (-6,6) * \frac{1}{3}[/tex]

[tex]A' = (-2,2)[/tex]

[tex]B' = (6,6) * \frac{1}{3}[/tex]

[tex]B' = (2,2)[/tex]

[tex]C' = (6,-6) * \frac{1}{3}[/tex]

[tex]C' = (2,-2)[/tex]