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vladimir2022 [97]

3 years ago

5

Write the formula for the area of a triangle, A, where b is the base and h is the height of the triangle. Don’t include parenthe

ses in your answer

1 answer:

stich3 [128]3 years ago

4 0

Area= (1/2 b) * h needs l=to hit character limit

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Solve by substitution <br> X^2= 2y + 10<br> 3x-y = 9

Viefleur [7K]

Answer:

$x=-4\\ x=-2\\ y=-21\\ y=-15$

Step-by-step explanation:

$x^{2} =2y+10$

$y=-9+3x$

$x^{2} =2(-9+3x)+10$

$x^{2} =-18+6x+10$

$x^{2} =6x-8$

$x^{2} -6x+8$

$x=-4, -2$

$y=-9+3(-4)$

$y=-9-12$

$y=-21$

$y=-9+3(-2)$

$y=-9-6$

$y=-15$

8 0

3 years ago

Read 2 more answers

Find a negative integer M so that the equation (x-12)(x+m) = m-51

andriy [413]

Answer:

dsds

Step-by-step explanation:

dwadwdadw

3 0

2 years ago

Choose the two triangles that are similar

levacccp [35]

Answer:triangles 20 12 28 and15 9 21

Step-by-step explanation:

6 0

2 years ago

Don buys 6 kiwis for 3$ what would a customer have to pay for 9 kiwis

kirill [66]

Answer:

It will cost $4.50 for 9 kiwis

Step-by-step explanation:

Lets find the cost per kiwi

$3 / 6 kiwis

.50 per kiwi

Now we have 9 kiwis

$.50 (9)

$4.50

It will cost $4.50 for 9 kiwis

3 0

2 years ago

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Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.

sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

All the sides of the rhombus are congruent: $|DG|\cong |GF| \cong |EF| \cong |DE|$
The diagonals are perpendicular

Using the distance formula; $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}$

$\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}$

$|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}$

$\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}$

$|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}$

$\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}$

$|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}$

$\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}$

Using the slope formula; $m=\frac{y_2-y_1}{x_2-x_1}$

The slope of EG is $m_{EG}=\frac{2c-0}{0-0}$

$\implies m_{EG}=\frac{2c}{0}$

The slope of EG is undefined hence it is a vertical line.

The slope of DF is $m_{DF}=\frac{c-c}{a+b-(-a-b)}$

$\implies m_{DF}=\frac{0}{2a+2b)}=0$

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since $|DG|\cong |GF| \cong |EF| \cong |DE|$ and $DF \perp FG$ , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is $m_{DE}=\frac{2c-c}{0-(-a-b)}$

$m_{DE}=\frac{c}{a+b)}$

The slope of FG is $m_{FG}=\frac{c-0}{a+b-0}$

$\implies m_{FG}=\frac{c}{a+b}$

5 0

3 years ago