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3 years ago

Answer number 1 and 2 am giving brainliest please help

Mathematics

2 answers:

Ymorist [56]3 years ago

5 0

9514 1404 393

Answer:

38 square inches

Step-by-step explanation:

The figure is a quadrilateral with one pair of parallel sides. That means it is a trapezoid, so the area formula for a trapezoid is applicable.

A = (1/2)(b1 +b2)h

The two parallel bases are 7 in and 12 in. The height is the perpendicular distance between them, 4 in. So, the area is ...

A = (1/2)(7 in +12 in)(4 in) = (1/2)(19 in)(4 in)

A = 38 in²

The area of the trapezoid is 38 square inches.

natali 33 [55]3 years ago

5 0

Answer:

<h2><u>Answer </u><u>:</u><u>-</u></h2>

We know that

The shape is trapezoid.

According to Question,

The area of trapezoid = ½(a + b) × h

a and b are parallel side of the trapezoid
H is the height of trapezoid

$\large \sf \: \dfrac{1}{2} \bigg(12 + 7 \bigg) \times 4$

$\large \sf \: 1 \times 19 \times 2$

$\large \sf \: 19 \times 2$

$\large \sf \: 38 \: in {}^{2}$

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Mrrafil [7]

see the correct question in the attached file

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3 years ago

Evaluate 6 - 2(-1) + l -5 l =

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6 0

3 years ago

19) Albert says that the two systems of equations shown have the same solutions.

k0ka [10]

Answer:

option A) Agree, because the solutions are the same is correct.

Step-by-step explanation:

FIRST SYSTEM

$6x + y= 2$

$-x-y=-3$

solving the system

$\begin{bmatrix}6x+y=2\\ -x-y=-3\end{bmatrix}$

$\mathrm{Multiply\:}-x-y=-3\mathrm{\:by\:}6\:\mathrm{:}\:\quad \:-6x-6y=-18$

$\begin{bmatrix}6x+y=2\\ -6x-6y=-18\end{bmatrix}$

adding the equation

$-6x-6y=-18$

$+$

$\underline{6x+y=2}$

$-5y=-16$

so the system becomes

$\begin{bmatrix}6x+y=2\\ -5y=-16\end{bmatrix}$

solve -5y for y

$-5y=-16$

Divide both sides by -5

$\frac{-5y}{-5}=\frac{-16}{-5}$

simplify

$y=\frac{16}{5}$

$\mathrm{For\:}6x+y=2\mathrm{\:plug\:in\:}y=\frac{16}{5}$

$6x+\frac{16}{5}=2$

subtract 16/5 from both sides

$6x+\frac{16}{5}-\frac{16}{5}=2-\frac{16}{5}$

$6x=-\frac{6}{5}$

Divide both sides by 6

$\frac{6x}{6}=\frac{-\frac{6}{5}}{6}$

$x=-\frac{1}{5}$

Therefore, the solution to the FIRST SYSTEM is:

$x=-\frac{1}{5},\:y=\frac{16}{5}$

SECOND SYSTEM

$2x-3y = -10$

$-x-y=-3$

solving the system

$\begin{bmatrix}2x-3y=-10\\ -x-y=-3\end{bmatrix}$

$\mathrm{Multiply\:}-x-y=-3\mathrm{\:by\:}2\:\mathrm{:}\:\quad \:-2x-2y=-6$

$\begin{bmatrix}2x-3y=-10\\ -2x-2y=-6\end{bmatrix}$

$-2x-2y=-6$

$+$

$\underline{2x-3y=-10}$

$-5y=-16$

so the system of equations becomes

$\begin{bmatrix}2x-3y=-10\\ -5y=-16\end{bmatrix}$

solve -5y for y

$-5y=-16$

Divide both sides by -5

$\frac{-5y}{-5}=\frac{-16}{-5}$

Simplify

$y=\frac{16}{5}$

$\mathrm{For\:}2x-3y=-10\mathrm{\:plug\:in\:}y=\frac{16}{5}$

$2x-3\cdot \frac{16}{5}=-10$

$2x=-\frac{2}{5}$

Divide both sides by 2

$\frac{2x}{2}=\frac{-\frac{2}{5}}{2}$

Simplify

$x=-\frac{1}{5}$

Therefore, the solution to the SECOND SYSTEM is:

$x=-\frac{1}{5},\:y=\frac{16}{5}$

Conclusion:

As both systems of equations have the same solution.

Therefore, we conclude that Albert is right when says that the two systems of equations shown have the same solutions.

Hence, option A) Agree, because the solutions are the same is correct.

8 0

3 years ago