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2 years ago

Which order pair is a solution to the system of linear equations -4+y=8 and x-5y=17?

Mathematics

2 answers:

nata0808 [166]2 years ago

7 0

Answer:

(77,12)

Step-by-step explanation:

$-4+y=8$

$x-5y=17$

We need to solve for x and y

USe substitution method, we solve the first equation for y

$-4+y=8$ , add 4 on both sides to get y alone

$y=12$

Now we plug in 12 for y in the second equation and find out x

$x-5y=17$

$x-5(12)=17$

$x-60=17$

Add 60 on both sides

$x= 77$

The ordered pair is (x,y). our x=77 and y=12

So ordered pair is (77,12)

andreev551 [17]2 years ago

4 0

Hello!

To find this ordered pair, we solve for x and y and put them together in an ordered pair.

-4+y=8

x-5y=17

Let's solve the first equation for y.

y= 12

Now, let's plug 12 into the second equation for y.

x-5(12)=17

x-60=17

x=77

Therefore, our ordered pair is (77,12)

I hope this helps!

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As given in the question,

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5. What is the solution to the equation -3(h + 5) + 2 = 4(h + 6) - 9

Klio2033 [76]

Answer: d.h=−4

PLZ MARK BRAINLIEST!

Step-by-step explanation:

Let's solve your equation step-by-step.

−3(h+5)+2=4(h+6)−9

Step 1: Simplify both sides of the equation.

−3(h+5)+2=4(h+6)−9

(−3)(h)+(−3)(5)+2=(4)(h)+(4)(6)+−9(Distribute)

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(−3h)+(−15+2)=(4h)+(24+−9)(Combine Like Terms)

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−3h−13=4h+15

Step 2: Subtract 4h from both sides.

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Step 3: Add 13 to both sides.

−7h−13+13=15+13

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Step 4: Divide both sides by -7.

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The length of the shadow of a pole on level ground increases by 90m when the angle of elevation of the sun changes from 58° to 3

viva [34]

Answer:

119.45 m

Step-by-step explanation:

Given:

When angle of elevation of the sun changes from 58° to 36° the length of shadow of a pole increases by 90 m.

To find:

Length of pole = ?

Solution:

Kindly refer to the attached image.

$\triangle ABC$ represents the 1st angle of elevation of sun i.e. 58°

$\triangle ABD$ represents the 2nd angle of elevation of sun i.e. 36°

Change in shadow is represented by CD = 90 m

Let height of pole, AB = $h$ m

Let side BC = $x$ m

Now, let us apply tangent rules in $\triangle ABC, \triangle ABD$ one by one:

$tan\theta = \dfrac{Perpendicular}{Base}\\\Rightarrow tan58^\circ=\dfrac{AB}{BC}\\\Rightarrow tan58^\circ=\dfrac{h}{x}\\\Rightarrow x = 0.624h ..... (1)$

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Putting value of $x$ using equation (1):

$tan36^\circ = \dfrac{h}{0.624h+90}\\\Rightarrow 0.726\times 0.624h+0.726\times 90 = h\\\Rightarrow h-0.453h =65.34\\\Rightarrow \bold{h = 119.45\ m}$

119.45 m is the height of pole.

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