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3 years ago

Find the solution of this system of equations. x-8y=21 and 3x-8y=15

Mathematics

1 answer:

dusya [7]3 years ago

8 0

Answer:

x = -3 , y = -3

Step-by-step explanation:

Solve the following system:

{x - 8 y = 21 | (equation 1)

3 x - 8 y = 15 | (equation 2)

Swap equation 1 with equation 2:

{3 x - 8 y = 15 | (equation 1)

x - 8 y = 21 | (equation 2)

Subtract 1/3 × (equation 1) from equation 2:

{3 x - 8 y = 15 | (equation 1)

0 x - (16 y)/3 = 16 | (equation 2)

Multiply equation 2 by 3/16:

{3 x - 8 y = 15 | (equation 1)

0 x - y = 3 | (equation 2)

Multiply equation 2 by -1:

{3 x - 8 y = 15 | (equation 1)

0 x+y = -3 | (equation 2)

Add 8 × (equation 2) to equation 1:

{3 x+0 y = -9 | (equation 1)

0 x+y = -3 | (equation 2)

Divide equation 1 by 3:

{x+0 y = -3 | (equation 1)

0 x+y = -3 | (equation 2)

Collect results:

Answer: {x = -3 , y = -3

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A line passes through (−1, 7) and (2, 10).

Pachacha [2.7K]

The slope m of a line through points $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by :

$\displaystyle{m = \frac{y_2-y_1}{x_2-x_1}
$

Thus, the slope of the line passing through points (−1, 7) and (2, 10) is

 $\displaystyle{m= \frac{10-7}{2-(-1)}= \frac{3}{3}=1$


The equation of a line with slope m passing through a point P(a, b) is given by
(y-b)=m(x-a).

We can consider any of the points (-1, 7), or (2, 10). Let's choose (2, 10):

y-10=1(x-2)
y-10=x-2
y-x=-2+10
y-x=8

Answer: C. −x+y=8

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3 years ago

In ΔPQR, the measure of ∠R=90°, the measure of ∠P=26°, and PQ = 8.5 feet. Find the length of QR to the nearest tenth of a foot.

Vitek1552 [10]

Answer:

3.7feet

Step-by-step explanations

Using the sin rule

A/sin a = B/sin b

Let A = PQ = 8.5feet

B = QR = x feet

a = R = 90°

b = P = 26°

Substitute the values into the Sin rule

8.5/sin90 = x/sin26

8.5×sin 26 = x × sin 90

8.5×0.4383 = x× 1

3.7255 = x

Hence the length of QR to the nearest tenth 3.7feet

5 0

2 years ago

Read 2 more answers

Hellp iv been trying for daysssssss

Shkiper50 [21]

Answer:

There are mathematics you could purchase that has answer sheets, what grade you in?

5 0

2 years ago

How do you find the area of the shaded region

Amiraneli [1.4K]

well, first off, let's notice that we have a trapezoid with a rectangle inside it, so the rectangle is really "using up" area that the trapezoid already has.

now, if we just get the area of the trapezoid, and then the area of the rectangle alone, and then subtract that area of the rectangle, the rectangle will in effect be making a hole inside the trapezoid's area, and what's leftover, is the shaded section, that part the hole is not touching.

$\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h&=height\\ a,b&=\stackrel{parallel~sides}{bases} \\\cline{1-2} h&=8\\ a&=16\\ b&=8 \end{cases}\implies A=\cfrac{8(16+8)}{2}\implies A=96 \\\\\\ \stackrel{\textit{area of the rectangle}}{(5\cdot 3)\implies 15}~\hfill \stackrel{~\hfill \textit{shaded area}}{\stackrel{\textit{trapezoid}}{96}~~ - ~~\stackrel{\textit{rectangle}}{15}~~ = ~~81}$

3 0

3 years ago

1. The shape shown below is dilated by a scale factor of 2.6. It is in the attachment that was included.

kondaur [170]

Answer:

L' = 26cm

W' = 39cm

Step-by-step explanation:

a) Given the dimension of the shape;

length L = 10cm

Width W = 15cm

After dilating by a scale factor os 2.6

New length L' = 2.6(10) = 26cm

New width = W' = 2.6(15) = 39cm

Hence the length of each sides of the dilated image are 26cm and 39cm

b) The new image will be similar to the given image but of with the resulting length gotten in (a).

7 0

2 years ago