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Lena [83]
4 years ago
10

Is 2750 a reasonable answer for 917 x 33?

Mathematics
1 answer:
irinina [24]4 years ago
7 0
Line them up on top of each other so it would be 30,261 so you were close but you have to line them up correctly
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A kitchen measures 3.75 meters by 4.2 meters.
dalvyx [7]

Answer:

a. A_{(kitchen)}=15.75\ m^2

b. A_{(total)}=39.375\ m^2

Step-by-step explanation:

The formula for calculate the area of a rectangle is:

A=lw

Where "l" is the lenght and "w" is the width

a. We know that the kitchen measures 3.75 meters by 4.2 meters; then we can say that:

l=3.75\ m\\\\w=4.2\ m

Therefore, substituting these values into the formula, we get that the area of the kitchen is:

A_{(kitchen)}=(3.75\ m)(4.2\ m)=15.75\ m^2

b. The area of the living room is one and a half times (1.5 times) that of the kitchen, this is:

A_{(living)}=(1.5)(15.75\ m)=23.625\ m^2

Therefore, the total of the living room and the kitchen is:

A_{(total)}=15.75\ m^2+23.625\ m^2=39.375\ m^2

3 0
3 years ago
How is the process of this operation? <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Csqrt%5B6%5D%7Bz%2B5%7D%20%7D%7B%20%5C
Yuki888 [10]
a^\frac{n}{m}=\sqrt[m]{a^n}\\---------------\\\\ \dfrac{ \sqrt[6]{z+5} }{ \sqrt{z+5} } =(z+5)^\frac{1}{6}:(z+5)^\frac{1}{2}\\\\/use:a^n\cdot a^m=a^{n+m}/\\\\=(z+5)^{\frac{1}{6}+\frac{1}{2}}=(z+5)^{\frac{1}{6}+\frac{3}{6}}=(z+5)^\frac{4}{6}=(z+5)^\frac{2}{3}\\\\=\sqrt[3]{(z+5)^2}=\sqrt[3]{z^2+2z\cdot5+5^2}=\sqrt[3]{z^2+10z+25}
4 0
3 years ago
Suppose that a function​ f(x) is defined for all real values of x except at xequals=c. can anything be said about the existence
Margaret [11]

we are given that

f(x) is defined for all values of x except at x=c

Limit may or may not exist

case-1:

If there is hole at x=c , then limit exist

case-2:

If there is vertical asymptote at x=c , then limit does not exist

Examples:

case-1:

\lim_{x \to c} \frac{x^2-cx}{(x-c)}

We can simplify it

\lim_{x \to c} \frac{x(x-c)}{(x-c)}

=\lim_{x \to c} x

=c

so, we can see that limit exist and it's value defined

case-2:

\lim_{x \to c} \frac{1}{(x-c)}

Left limit is

\lim_{x \to c-} \frac{1}{(x-c)}

=-\infty

Right Limit is

\lim_{x \to c+} \frac{1}{(x-c)}

=+\infty

so, we can see that left limit is not equal to right limit

so, limit does not exist

4 0
3 years ago
What is the area of the figure above ?
RideAnS [48]

Answer:

256m^2

Step-by-step explanation:

you can cut up the figure.

A1=bh

A1= 24x8

A1=192m^2

A2=bh

A2= bh

A2=8x8

A2=64m^2

A1+A2= Atotal

A=192+64

A=256m^2

4 0
3 years ago
I am thinking of a number. If you add 3 to the number and then multiply the sum by four, you end up with 48. What is my number?
kap26 [50]

Answer:

9

Step-by-step explanation:

48 / 4 = 12

12 - 3 = 9

Check Work

9 + 3 = 12

12 * 4 = 48

5 0
3 years ago
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