All categories

2 years ago

PLEASE HELP I WILL MARK BRAINLIEST

Mathematics

1 answer:

Gnom [1K]2 years ago

3 0

Answer:

is C 4,816

Step-by-step explanation:

four thousand is 4,000 eight hundred is 800 sixteen 16

4000+800+16=4,816

You might be interested in

What is the domain of the function s=-4r^2 if the range is s <_ 0

vlada-n [284]

Answer:

r≤0

Step-by-step explanation:

if the range is s≤0, (that means the value of s is all negative integers &0) to find the domain we have to replace that some negative integers &0.

we can replace some values of s as the following

s=‒4(0)^2 s=‒4(‒1)^2

=‒4(0) =‒4(1)

=0 =‒4

s=‒4(‒2)^2 s=‒4(‒3)^2

=‒4(4) =‒4(9)

=‒16 =‒36

so we can understand easly that the domain is negative integer&0,it can't be positive because the square of any integer is positive.

therefore the answer is r≤0

i think &i hope it is clear.

3 0

2 years ago

Rectangle A is a scale drawing of Rectangle B and has 25% of its area. If rectangle A has side lengths of 4 cm and 5 cm , what a

SpyIntel [72]

Answer:

8 cm and 10 cm

Step-by-step explanation:

Hello, <em> </em>I can help you with this.

Step 1

According to the question there are two rectangles A and B,

Rectangle A is a scale drawing of Rectangle B and has 25% of its area

in other words

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\$

Step 2

Let

Rectangle A

length (1)= 4 cm

length (2)= 5 cm

$Area_{A}=4\ cm * 5\ cm\\Area_{A}=20\ cm^{2}$

put this value into equation 1

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\\\20\ cm^{2} =0.25*Area_{B} \\divide\ each\ side\ by\ 0.25\\\frac{20\ cm^{2} }{0.25}=\frac{0.25}{0.25}*Area_{B}\\ Area_{B}=80\ cm^{2}$

Now, we know the area of rectangle B, to know its length we need to formule other equation

Step 3

$Area_{B}=80\ cm^{2}\\length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\$

the ratio between the lengths must be constant, so the ratio of A must be equal to ratio in B, then

$\frac{length(1A)}{length(2A)}=\frac{length(1B)}{length(2B)} \\\\\\frac{4}{5}= \frac{length(1B)}{length(2B)}\\0.8=\frac{length(1B)}{length(2B)}\\length(1B)=0.8*length(2B) (Equation 3)$

Step three

using Eq 1 and Eq 2 find the lengths

put the value of length(1B) into equation (2)

$length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\\(0.8*length(2B)) (*length (2B)=80\ cm^{2} \\\\0.8*(length (2B))^{2} =80\ cm^{2}\\(length (2B))^{2} =\frac{80\ cm^{2}}{0.8} \\(length (2B))^{2}=100\\\sqrt{(length (2B))^{2}}=\sqrt{100\ cm^{2}} \\ length (2B)=10\ cm$