HELP ME ASAP on these math question it’s due after class

How do u do 6x32 and 7x29 using distribuative property to solve

Rufina [12.5K]

Answer:

use multiple the numbers to get the co rrect

Step-by-step explanation:

32×6

=192

29×7=203

4 0

2 years ago

Read 2 more answers

3 ^(1 − 7 x = 7 ^x find the exact value

Shalnov [3]

Answer:

x

=

ln

(

3

)

7

ln

(

3

)

+

ln

(

7

)

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

skad [1K]

Answer:

2

Step-by-step explanation:

2+2=2 and then 1 come now you have 1 2 and 1 1

5 0

2 years ago

Write a linear equation for slope: 0; y-intercept: -8

Novosadov [1.4K]

Answer:

x = -8

Step-by-step explanation:

Since there is no slope, there is no need for a y variable to appear in your final equation. The graph will cross y = -8 horizontally. So therefore, x = -8

This is a pretty short explanation, but there's not too much to it. Hope this helped :)

5 0

2 years ago

For the figures below, assume they are made of semicircles, quarter circles and squares. For each shape, find the area and perim

ICE Princess25 [194]

Answer:

Part a) The area of the figure is $\frac{9}{2}(4+\pi )\ cm^{2}$

Part b) The perimeter of the figure is $3(2+2\sqrt{2}+ \pi)\ cm$

Step-by-step explanation:

Step 1

Find the area of the figure

In this problem we have that

The figure ABC is the half of a square and the other figure is a semicircle

<u>Find the area of the figure ABC</u>

we have

$AB=6\ cm, BC=6\ cm$

The area of the half square ABC is equal to find the area of triangle ABC

so

$A1=\frac{1}{2}*6*6=18\ cm^{2}$

<u>Find the area of the semicircle</u>

The area of the semicircle is equal to

$A2=\pi r^{2}/2$

we have that

$r=6/2=3\ cm$

substitute

$A2=\pi (3)^{2}/2$

$A2=(9/2) \pi\ cm^{2}$

The area of the figure is equal to

$18\ cm^{2}+(9/2) \pi\ cm^{2}= \frac{9}{2}(4+\pi )\ cm^{2}$

Step 2

Find the perimeter of the figure

The perimeter of the figure is equal to

$P=AB+AC+length\ CB$

we have

$AB=6\ cm$

Applying Pythagoras theorem

$AC=\sqrt{6^{2}+6^{2}}\\AC=6\sqrt{2}\ cm$

Remember that

the circumference of a semicircle is equal to

$C=\frac{1}{2}2\pi r=\pi r$

$r=6/2=3\ cm$

$C=\pi(3)$

$C=3 \pi\ cm$

The perimeter of the figure is equal to

$P=6\ cm+6\sqrt{2}\ cm+3 \pi\ cm$

Simplify

$P=3(2+2\sqrt{2}+ \pi)\ cm$

5 0

3 years ago