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3 years ago

Write the decimal for 1/8. Explain why this decimal is called a terminating decimal

Mathematics

1 answer:

iren [92.7K]3 years ago

4 0

Answer:

0.125

Step-by-step explanation:

A terminating decimal has a set or non infinite amount of numbers after the decimal point.

this decimal is a terminating decimal simply because it has an end to it.

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How can 5/16be written as a decimal

juin [17]

Answer:

5/16 as a decimal is 0.3125.

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2 years ago

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The two right triangular prisms are similar solids.

Zanzabum

Answer:

<h2>The volume of the larger prims is eight times greater than the smaller prism.</h2>

Step-by-step explanation:

If the scale factor of the larger prims to the smaller prism is 3, that means each side is triple.

We know the a triangular prims has a volume defined by

$V_{small} =\frac{b \times h_{1} }{2} \times h_{2}$

If we take that formula as the smaller prism, then the larer prism has a volum of

$V_{large} =\frac{2b \times 2h_{1} }{2} \times 2h_{2}=8(\frac{b \times h_{1} \times h_{2} }{2} )$

If we compare, we can deduct that

$V_{large}=8V_{small}$

In words, the volume of the larger prims is eight times greater than the smaller prism.

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3 years ago

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Help me please with this maths homework

IRINA_888 [86]

The value of the number can be equal to or less than -3

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3 years ago

Select the letter that correctly identifies the base of the prism.

Radda [10]

Answer:

I think it's C, I may be wrong though

Step-by-step explanation:

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3 years ago

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|. Identify the following Pōints of each values.Write your ans

Dmitry_Shevchenko [17]

<h2>✒️VALUE</h2>

$\\ \quad \begin{array}{c} \qquad \bold{Distance \: \green{ Formula:}}\qquad\\ \\ \boldsymbol{ \tt d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}} \end{array}\\ \begin{array}{l} \\ 1.)\: \bold{Given:}\: \begin{cases}\tt D(- 5,6), E(2.-1),\textsf{ and }F(x,0) \\ \tt DF = EF \end{cases} \\ \\ \qquad\bold{Required:}\:\textsf{ value of }x \\ \\ \qquad \textsf{Solving for }x, \\ \\ \tt \qquad DF = EF \\ \\ \implies\small \tt{\sqrt{(x -(- 5))^2 + (0 - 6)^2} = \sqrt{(x - 2)^2 + (0 - (-1))^2}} \\ \\ \implies\tt\sqrt{(x + 5)^2 + 36 } = \sqrt{(x - 2)^2 + 1 } \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies\tt (x + 5)^2 + 36 = (x - 2)^2 + 1 \\ \\ \implies\tt x^2 + 10x + 25 + 36 = x^2 - 4x + 4 + 1 \\ \\ \implies \tt x^2 + 10x + 61 = x^2 - 4x + 5 \\ \\ \implies\tt10x + 4x = 5 - 61 \\ \\ \implies\tt14x = -56 \\ \\ \implies \red{\boxed{\tt x = -4}}\end{array} \\ \\ \\ \\\begin{array}{l} \\ 2.)\: \bold{Given:}\: \begin{cases}\tt P(6,-1), Q(-4,-3),\textsf{ and }R(0,y) \\ \tt PR = QR \end{cases} \\ \\ \bold{Required:}\:\textsf{ value of }y \\ \\ \qquad\textsf{Solving for }y, \\ \\ \qquad\tt PR = QR \\ \\ \implies \tt\small{\sqrt{(0 - 6)^2 + (y - (-1))^2} = \sqrt{(0 - (-4))^2 + (y - (-3))^2}} \\ \\ \implies\tt\sqrt{36 + (y + 1)^2} = \sqrt{16 + (y + 3)^2 } \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies \tt \: 36 + (y + 1)^2 = 16 + (y + 3)^2 \\ \\ \implies\tt 36 + y^2 + 2y + 1 = 16 + y^2 + 6y + 9 \\ \\ \implies \tt \: y^2 + 2y + 37 = y^2 + 6y + 25 \\ \\ \implies \tt \: 2y - 6y = 25 - 37 \\ \\ \implies \tt -4y = -12 \\ \\ \implies\red{\boxed{ \tt y = 3}} \end{array} \\ \\ \\ \begin{array}{l} \\ 3.)\: \bold{Given:}\: \begin{cases}\: A(4,5), B(-3,2),\textsf{ and }C(x,0) \\ \: AC = BC \end{cases} \\ \\ \bold{Required:}\:\textsf{ value of }x \\ \\ \qquad\textsf{Solving for }x, \\ \\ \qquad\tt AC = BC \\ \\ \implies\tt\small{\sqrt{(x - 4)^2 + (0 - 5)^2} = \sqrt{(x - (-3))^2 + (0 - 2)^2}} \\ \\ \implies\tt\sqrt{(x - 4)^2 + 25} = \sqrt{(x + 3)^2 + 4} \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies\tt\:(x - 4)^2 + 25 = (x + 3)^2 + 4 \\ \\ \implies\tt\:x^2 - 8x + 16 + 25 = x^2 + 6x + 9 + 4 \\ \\ \implies\tt\:x^2 - 8x + 41 = x^2 + 6x + 13 \\ \\ \implies\tt-8x - 6x = 13 - 41 \\ \\\implies\tt -14x = -28 \\ \\ \implies\red{\boxed{\tt\:x = 2}} \end{array}$

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2 years ago