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3 years ago

Yong is measuring two pyramids whose bases are rectangles.

Mathematics

1 answer:

ElenaW [278]3 years ago

7 0

Answer:

Answer is 7/2 cm

Step-by-step explanation:

1. Set a ratio

$7=\frac{3V}{l*w}:h=\frac{3V}{l*2w}$

2. Move 2w from denominator to LHS (Left hand side) $\frac{7}{2}w=\frac{3V}{l}$

$\frac{7}{2w}=\frac{3V}{l}$

3. The four options are 7/4, 7/2, 14 or 28 cm

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Can anyone help me with my homework please

Over [174]

1..... 8(x+6)=-64
8x+48 =-64
8x =-64-48

8x=-112
X=-112/8
x=-14....

2.....3(4-2y)=-15
12-6y=-15

-6y=-15-12

-6y = - 27
y=27/6

y=4½....

3.....8w-3w+6=41
5w=41-6
5w=35
W=35/5
w=7....

4.....9x+3+x=-77

10x=-77-3
10x=-80
x=-8...

5.....10(-x-2)=120

-10x-20=120

-10x=120+20
-10x=140
x=140/-10
x=-14...

You shouldn't have a problem with the rest now....

7 0

2 years ago

Read 2 more answers

For the rectangle shown, which equation can be used to find the value of x?

trasher [3.6K]

Answer:

Pythagorean theorem $a^{2} +b^{2} =c^{2}$

Step-by-step explanation:

$a^{2} +b^{2} =c^{2} \\x^{2} +3.6^{2} =7.4^{2} \\x^{2} +12.96=54.76\\x^{2} =\sqrt{41.8} \\x=6.47\\$

6 0

2 years ago

Long division<br> 3x+1/6x^6+5x^5+2x^4-9x^3+7x^2-10x+2

inna [77]

$(6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2) / (3x+1)$

We divide first number from first parenthesis with first number from second parenthesis. Then the resulting number we multiply by all numbers in second parenthesis and substract from first parenthesis.

$6 x^{6}/3x = 2 x^{5} \\ \\ 6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2 - 6 x^{6} -2 x^{5} = 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2\\$

We repeat previous steps until we run out of numbers:
$3x^{5}/3x=x^{4} \\ \\ 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2-3x^{5}-x^{4}= \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2 \\ \\ \\ x^{4}/3x= \frac{1}{3} x^{3} \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2-x^{4}- \frac{1}{3} x^{3}= \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2 \\ \\ \\ - \frac{28}{3} x^{3}/3x= - \frac{28}{9} x^{2} \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2+ \frac{28}{3} x^{3}+ \frac{28}{9} x^{2} = \\ \\ \frac{91}{9} x^{2}-10x+2$
$\frac{91}{9} x^{2}/3x=\frac{91}{27} x \\ \\ \frac{91}{9} x^{2}-10x+2-\frac{91}{9} x^{2}-\frac{91}{27} x= \\ \\ -\frac{361}{27} x+2 \\ \\ \\ -\frac{361}{27} x/3x=-\frac{361}{81} \\ \\ -\frac{361}{27} x+2+\frac{361}{27}x+\frac{361}{27}= \\ \\ \frac{415}{27}$

We are left with a number that has no x inside. This is remainder.
The final solution is sum of all these solutions and remainder:
$(2 x^{5}+x^{4}+\frac{1}{3} x^{3} - \frac{28}{9} x^{2} +\frac{91}{27} x)+(-\frac{361}{81} )$

6 0

2 years ago

What is 4/5 out of 60

CaHeK987 [17]

4/5 of 60 = 48. 4/5 = .80 Multiply .80 by 60 =48

8 0

3 years ago

Other ways to write 50897

NARA [144]

Written form: fifty thousand eight hundred ninety-seven.

expanded form: 50000+800+90+7

3 0

2 years ago

Read 2 more answers