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Ganezh [65]
3 years ago
14

Find the width (w) of a rectangular room with a length of 60 meters and an area (a) of 2,400 square meters

Mathematics
1 answer:
Olenka [21]3 years ago
3 0

Answer:

Romeo Whigh is 40m.

Step-by-step explanation:

P=a*b,

a=60m, b=wight(w), P=2400m^2

w=?

Change known size in P=a*b

2400=60*w

w=2400/60

w=40m

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Find the surface area of the prism.
stira [4]

Answer:

386

Step-by-step explanation:

6 0
3 years ago
Find fifth roots of z+32= 0 and draw its graph
madreJ [45]

Answer:

\sqrt[5]{z}  = \sqrt[5]{(-2)^{5} }   \therefore \sqrt[5]{z} = -2

Step-by-step explanation:

i) z = -32

ii) \sqrt[5]{z}  = \sqrt[5]{(-2)^{5} }   \therefore \sqrt[5]{z} = -2

8 0
2 years ago
Which division expression is equivalent to 4 1/3 ÷ -5/6
Rzqust [24]

4\dfrac{1}{3}\div\left(-\dfrac{5}{6}\right)\qquad(*)\\\\\text{Convert the mixed number to the improper fraction:}\\\\4\dfrac{1}{3}=\dfrac{4\cdot3+1}{3}=\dfrac{12+1}{3}=\dfrac{13}{3}\\\\(*)=\dfrac{13}{3}\div\left(-\dfrac{5}{6}\right)\\\\\text{By dividing by a fraction, we multiply by its reciprocal.}\\\\=\dfrac{13}{3}\cdot\left(-\dfrac{6}{5}\right)\qquad\text{simplify}\\\\=\dfrac{13}{3\!\!\!\!\diagup_1}\cdot\left(-\dfrac{6\!\!\!\!\diagup^2}{5}\right)=-\dfrac{13\cdot2}{1\cdot5}=-\dfrac{26}{5}=-\dfrac{25+1}{5}=-5\dfrac{1}{5}

5 0
3 years ago
Read 2 more answers
Make a table of values for -3 < x > 3 and plot a graph for y= -2x + 3
denis-greek [22]
X= -3x + 2 this is the answer

4 0
3 years ago
Consider m = y2 - y1/ x2 - x1 . Which x1 and x2-values would determine that the line is vertical? Justify your answer
Y_Kistochka [10]

Answer:

x_2=x_1

Step-by-step explanation:

We were given the slope formula;

m=\frac{y_2-y_1}{x_2-x_1}

This line is vertical if the denominator is zero.

That is when x_2-x-1=0

This implies that;

x_2=x_1

Justification;

When x_2=x_1, then, the line passes through;

(x_1,y_1)  and (x_1,y_2)

The slope now become

m=\frac{y_2-y_1}{x_1-x_1}=\frac{y_2-y_1}{0}

The equation of the line is

y-y_1=\frac{y_2-y_1}{0}(x-x_1)

This implies that;

0(y-y_1)=(y_2-y_1)(x-x_1)

0=(y_2-y_1)(x-x_1)

\frac{0}{y_2-y_1}=(x-x_1)

0=(x-x_1)

x=x_1... This is the equation of a vertical line.

3 0
2 years ago
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