Help!!!!!!! please !!!!!!!

Mathematics

2 answers:

Fynjy0 [20]3 years ago

7 0

C is the answer hope it helps

Soloha48 [4]3 years ago

4 0

The corners are right angles, but the sides are different lengths. The figure is a rectangle.

You might be interested in

How do you solve this problem?

babymother [125]

Let's solve your equation step-by-step.16(a−4)=13(2a+4)Step 1: Simplify both sides of the equation.16(a−4)=13(2a+4)(16)(a)+(16)(−4)=(13)(2a)+(13)(4)(Distribute)16a+−23=23a+43Step 2: Subtract 2/3a from both sides.16a+−23−23a=23a+43−23a−12a+−23=43Step 3: Add 2/3 to both sides.−12a+−23+23=43+23−12a=2Step 4: Multiply both sides by 2/(-1).(2−1)*(−12a)=(2−1)*(2)a=−4

7 0

3 years ago

Read 2 more answers

Given the points A(-1,-3) and B(5,2). A. Write the equation of the straight line AB. Another line MN is parallel to AB and passe

Alexus [3.1K]

Find slope of AB where A = (-1,-3) and B = (5,2):

$\text {Slope = } \dfrac{Y_2-Y_1}{X_2-X_1}$

$\text {Slope = } \dfrac{2+3}{5+1} = \dfrac{5}{6}$

Write equation with the slope found:

$y = mx + b$

$y = \dfrac{5}{6} x + b$

Find y-intercept:

$y = \dfrac{5}{6} x + b$

$2 = \dfrac{5}{6} (5) + b$

$2 = \dfrac{25}{6} + b$

$b = 2 - \dfrac{25}{6}$

$b = -\dfrac{13}{6}$

Form the equation:

$y = \dfrac{5}{6} x - \dfrac{13}{6}$

$\boxed {\boxed {\bf \text {Answer: Line AB : }y = \dfrac{5}{6} x - \dfrac{13}{6}}}$


Find Slope of Line MN where MN is parallel to AB:

$\text {Slope of MN = Slope of AB = } \dfrac{5}{6}$

Write equation with the slope found:

$y = mx + b$

$y = \dfrac{5}{6} x + b$

Find y-intercept given that the line passed through (6, -2):

$y = \dfrac{5}{6} x + b$

$-2 = \dfrac{5}{6} (6) + b$

$-2 =5 + b$

$b = -2 - 5$

$b = -7$

$
\boxed {\boxed {\bf \text {Answer: Line MN } : y = \dfrac{5}{6} x -7 }}$

8 0

4 years ago

Mr. Howard is a student teacher at the local junior high. His first‐period class of 40 students averaged 96% on a recent test. H

kodGreya [7K]

Answer:

The combined average is 94%

Step-by-step explanation:

for first period, number of students =40

average =96%

for second period, no of students =20

average =90%

but $average = \frac{sum of scores}{no of students}$