All categories

4 years ago

Can anyone help me with this??

Mathematics

1 answer:

fredd [130]4 years ago

6 0

7)X=20
8)X=14
9)X=45.83 repeating
10)X=-19
11)X=100
12)X=-25

You might be interested in

Here is a system of equations.

goblinko [34]

Answer:

It should be "one solution"

Step-by-step explanation:

After graphing the equations, the two lines only intersect at one point which makes it "one solution." Hope this helps.

8 0

3 years ago

Read 2 more answers

Add the two expressions. m + 6 and 4m + 2 Enter your answer in the box

ANEK [815]

Hey so we would collect like terms: 5m+8

If this helped, please consider brainliest! Have an awesome day!

4 0

3 years ago

Read 2 more answers

Find the circumference of a circle whose area is equal to 64π^2

PilotLPTM [1.2K]

Answer:

14 6/7 is the circumference

Step-by-step explanation:

you first of all look for the radius using the formulae multiply both sides by the reciprocal of 22/7

7/22 ×22/7r=64×7/22=224/11{2 4/11)

2 4/11 is the radius

cir =2×22/7×2 4/11

2×22/7× 26/11=104/7

==== =14 6/7

all i can do but am not sure sorry

7 0

2 years ago

A cyclist travels at distance of 400 meters in 120 seconds towards school, calculate his speed. (Show your work)

DochEvi [55]

we know that

The scalar magnitude of the velocity vector is the speed. The speed is equal to

$Speed=\frac{distance}{time}$

in this problem we have

$distance=400\ m \\time=120\ sec$

substitute in the formula

$Speed=\frac{400}{120}$

$Speed=3.5\frac{m}{sec}$

therefore

the answer Part a) is

the speed is equal to $3.5\frac{m}{sec}$

Part b) Find the velocity

we know that

Velocity is a vector quantity; both magnitude and direction are needed to define it

in this problem we have

the magnitude is equal to the speed

$magnitude=3.5\frac{m}{sec}$

$direction=North\ East\ (NE)$

therefore

the answer Part b) is

the velocity is $3.5\frac{m}{sec}\ North\ East\ (NE)$

Part c)

we know that

the acceleration is equal to the formula

$a=\frac{V2-V1}{t2-t1}$

in this problem we have

$V2=0 \\V1=3.5\frac{m}{sec}$

$t2=15\ sec\\t1=0$

substitute in the formula

$a=\frac{0-3.5}{15-0}$

$a=-\frac{3.5}{15}\frac{m}{sec^{2}}$

$a=-\frac{7}{30}\frac{m}{sec^{2}}$

therefore

the answer Part c) is

the acceleration is $-\frac{7}{30}\frac{m}{sec^{2}}$

This is an example of negative acceleration

7 0

3 years ago

Read 2 more answers

Select an equation that is equivalent to a^2-144b^2 a) (a-12b)^2 B) (a+12b)^2 C) (a-12b)(a+12b) C) a^2-24ab+144b^2

Nataliya [291]

Answer:

C) (a-12b)(a+12b)

Step-by-step explanation:

a^2-144b^2 = (a-12b)(a+12b)

This is the formula that when two terms squared are written with a minus sign between them their factors are the product of the original numbers once with a plus sign and once with the minus sign.

This can be explained by multiplying them.

(a-12b)

*(a+12b)

a²- 12ab

+12ab - 144b²

a² +0 - 144b²

We see we get the same answer as applying the formula above.

4 0

3 years ago