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3 years ago

Help plz I am bad at this type of math

Mathematics

2 answers:

S_A_V [24]3 years ago

6 0

Answer:

Step-by-step explanation:

rusak2 [61]3 years ago

3 0

Answer:

the answer is D

Step-by-step explanation:

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How do you solve this problem<br> 8a+a-3=6a-2a-3

Nadya [2.5K]

$8a+a-3=6a-2a-3\\\\9a-3=4a-3\ \ \ \ /+3\\\\9a-3+3=4a-3+3\\\\9a=4a\ \ \ \ /-4a\\\\9a-4a=4a-4a\\\\5a=0\ \ \ \ /:5\\\\5a:5=0:5\\\\a=0$

7 0

4 years ago

Read 2 more answers

NEED HELP ASAP!! WILL MARK BRAINLIEST IF ITS RIGHT!!

forsale [732]

Answer:

the answer is right there x.8.90.^.34>^

Step-by-step explanation:

7 0

4 years ago

BEST ANSWER GETS BRAINLIEST!! HELP PLEASE.

Goshia [24]

Answer:

$\text{The area of parallelogram is }8 ft^2$

$\text{The area of two triangles of equal area is }4 ft^2$

Step-by-step explanation:

Given a parallelogram ABCD is shown with DC equal to 9 feet and the perpendicular distance between AB and DC equal to 8 over 9 foot.

Part A: we have to find the area of ABCD.

$\text{Area of parallelogram=}Base\times height$

$=9\times \frac{8}{9}=8 ft^2$

$\text{The area of parallelogram is }8 ft^2$

Part B:

We have to decompose this parallelogram into two triangles.

Join diagonal AC

As the diagonal of parallelogram divides it into two congruent triangles of equal area

Hence, the area of each triangle is

$\frac{1}{2}ar(ABCD)$

$\frac{1}{2}\times 4=4 ft^2$

$\text{The area of two triangles of equal area is }4 ft^2$

3 0

3 years ago

Find the lengths of the sides of the triangle PQR. P(2, −3, −4), Q(8, 0, 2), R(11, −6, −4) |PQ| = Incorrect: Your answer is inco

Aloiza [94]

Answer:

the length PQ is 9 units,the length QR is 9 units,the length PR is 9.48 units,the triangle is not a right triangle,this is a isosceles triangle

Step-by-step explanation:

Hello, I think I can help you with this

If you know two points, the distance between then its given by:

$P1(x_{1},y_{1},z_{1} ) \\P2(x_{2},y_{2},z_{2})\\\\d=\sqrt{(x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2}+(z_{2}-z_{1} )^{2} }$

Step 1

use the formula to find the length PQ

Let

P1=P=P(2, −3, −4)

P2=Q=Q(8, 0, 2)

$d=\sqrt{(8-2)^{2} +(0-(-3))^{2}+(2-(-4))^{2}} \\ d=\sqrt{(6)^{2} +(3)^{2}+(6 )^{2}}} \\d=\sqrt{36+9+36}\\d=\sqrt{81} \\d=9\\$

the length PQ is 9 units

Step 2

use the formula to find the length QR

Let

P1=Q=Q(8, 0, 2)

P2=R= R(11, −6, −4)

$d=\sqrt{(11-8)^{2} +(6-0))^{2}+(-4-2 )^{2}} \\\\\\d=\sqrt{(3)^{2} +(6)^{2}+(-6 )^{2}}} \\d=\sqrt{9+36+36}\\d=\sqrt{81} \\d=9\\$

the length QR is 9 units

Step 3

use the formula to find the length PR

Let

P1=P(2, −3, −4)

P2=R= R(11, −6, −4)

$d=\sqrt{(11-2)^{2} +(-6-(-3)))^{2}+(-4-4 )^{2}} \\\\\\d=\sqrt{(9)^{2} +(-6+3)^{2}+(-4-(-4) )^{2}}} \\d=\sqrt{81+9+0}\\d=\sqrt{90} \\d=9.48\\$

the length PR is 9.48 units

Step 4

is it a right triangle?

you can check this by using:

$side^{2} +side^{2}=hypotenuse ^{2}$

Let

side 1=side 2= 9

hypotenuse = 9.48

Put the values into the equation

$9^{2} +9^{2} =9.48^{2}\\ 81+81=90\\162=90,false$

Hence, the triangle is not a right triangle

Step 5

is it an isosceles triangle?

In geometry, an isosceles triangle is a type of triangle that has two sides of equal length.

Now side PQ=QR, so this is a isosceles triangle

Have a great day

3 0

3 years ago

Which of the following lines is perpendicular to the line 8x + 6y = 12 and passes through the point (-12,-2)?

VLD [36.1K]

Answer:

y = 3/4x + 7

Step-by-step explanation:

8x + 6y = 12

6y = -8x + 12

y = -4/3x + 2

Perpendicular slope: 3/4

y - y1 = m(x - x1)

y + 2 = 3/4(x + 12)

y + 2 = 3/4x + 9

y = 3/4x + 7

5 0

3 years ago