All categories

3 years ago

Can someone plz help me with this one problem!!!!

Mathematics

2 answers:

Bess [88]3 years ago

4 0

Answer:

It's 80.

Step-by-step explanation:

Galina-37 [17]3 years ago

3 0

Answer:

Step-by-step explanation:

5 × 4 = 20

20 × 4 = 80

80 × 4 = 320

320 × 4 = 1280

You might be interested in

Sally has a GPA of 3.35. What percentage of the students have a GPA above her GPA? Show all work. (Note: GPA is normally distrib

DaniilM [7]

The distribution lies within one of the standard of deviation of the mean so <span>68% </span>
The distribution lies within two of the standard of deviations of the mean so 95%
The distribution lies within three of the standard of deviations of the mean so 99.7%

7 0

3 years ago

In the given right triangle, AX is the altitude. If CX-9 and AX - 9, what is the length of BC? "Note: Figure not drawn to scale"

kenny6666 [7]

Answer:

Step-by-step explanation:

I’m pretty sure the answer is C

6 0

2 years ago

Read 2 more answers

Find the missing side lengths. Leave your answers as radicals in simplest<br> form.

JulsSmile [24]

X=40
Y=20. Are the awnsers

8 0

3 years ago

Simplify . (1/c + 1/h)/(1/(c ^ 2) - 1/(r ^ 2))

Marrrta [24]

Answer:

$\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Step-by-step explanation:

$\frac{\frac{1}{c}+\frac{1}{h}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine $\frac{1}{c} + \frac{1}{h}$

$\frac{\frac{h+c}{ch}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine the bottom, too.

$=\frac{\frac{h+c}{ch}}{\frac{r^2-c^2}{c^2r^2}}$

Apply the fraction rule

$=\frac{\left(h+c\right)c^2r^2}{ch\left(r^2-c^2\right)}$

Cancel

$=\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Therefore, $\frac{\left(\frac{1}{c}+\frac{1}{h}\right)}{\left(\frac{1}{\left(c^2\right)}-\frac{1}{\left(r^2\right)}\right)}:\quad \frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

5 0

3 years ago

What is the inverse of the function f(x)= x +3

pshichka [43]

Hello,

if f(x)=y=x+3

x=y+3==>y=x-3

f^(-1) (x)=x-3 is the inverse function of f(x).

5 0

3 years ago