Ask question

Ask question

All categories

3 years ago

12

40 POINTS! HELP! ASAP!

1 answer:

Sindrei [870]3 years ago

7 0

What is the greatest common factor of the expression 63r^2t^3+42r^3t^5
Question 4 options:
10r^3t^5
21r^2t^3
7r^2t^3
3r^5t^8

Answer is 21r^2t^3

You might be interested in

Can someone help? Please

MrRa [10]

Answer: I can’t help with that because there should be a graph

Step-by-step explanation:

4 0

3 years ago

I really need help with this ?

Digiron [165]

The answer is g. 47.1

you use the formula
c= π d
then plug is 15 for d and you should get the answer

8 0

3 years ago

Solve the inequality. x - 1 is less than or equal to -9.

Marizza181 [45]

Answer:

the answer is -8

Step-by-step explanation:

x<=-8

the answer is -8

7 0

3 years ago

What is the domain of the absolute value function below?

vovikov84 [41]

The domain is all real numbers. (-infinity, infinity)
You can put any real number x value into the function.

5 0

3 years ago

Read 2 more answers

Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

3 0

2 years ago