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3 years ago

What is the absolute value of -190 ? What is the absolute value of 65 ?

Mathematics

2 answers:

lbvjy [14]3 years ago

6 0

190 and 65 because no negatives are allowed

avanturin [10]3 years ago

5 0

| - 190 | = 190

Finding the absolute value is taking away the - sign if it is negative and if it is positive keeping it how it is. The absolute value is the number, not the sign.

Finding the absolute value is actually very simple, just take away the negative sign or if it is positive keep it how it is

In this scenario - 190 is the number. Like it says above just take away the negative sign. That gives you 190, 190 is the answer.

| 65 | = 65

Like above if it is positive you keep it how it is.

So 65's absolute value is it's self. 65. 65 is the answer.

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On Friday Merinda uses 1/4 of the box of pancake mix to make pancakes. On Saturday she uses 1 1/2 times as much pancake mix as F

Mashutka [201]

Answer:

3/8 of the box

Step-by-step explanation:

On Friday Merinda uses 1/4 of the box of pancake mix to make pancakes.

On Saturday she uses 1 1/2 times as much pancake mix as Friday.

On Friday = 1/4 of the box

On Saturday = 1 1/2 × pancake mix as Friday

$\dfrac{3}{2}\times \dfrac{1}{4}\\\\=\dfrac{3}{8}$

Hence, 3/8 of the box of pancake mix does Merinda use on Saturday.

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3 years ago

the volume of a cylindrical watering can is 100cm^3. If the radius is doubled, then how much water can the new can hold?

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V=(pi)(2r^2)(h)
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3 years ago

4 = 32x

inn [45]

Answer:

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Step-by-step explanation:

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3 years ago

A garden is shaped like a rectangle whose perimeter is 150 ft. The length is 2 times as long as the width. Find the length and t

vlabodo [156]

Answer:

Step-by-step explanation:

L=2W

P=2(L+W), using L=2W this becomes

P=2(2W+W)

P=2(3W)

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6W=150

W=25, and since L=2W

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2 years ago

ASAP someone please simply two exponents

zhannawk [14.2K]

Answer:

1.) $m^{15}$

2.) $=\frac{1}{y^{15}}$

Give me a comment if you want the explanation.

1.) $\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$\left(m^3\right)^5=m^{3\cdot \:5}$

$\mathrm{Multiply\:the\:numbers:}\:3\cdot \:5=15$

$=m^{15}$

2.) $\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$\left(y^{-3}\right)^5=y^{-3\cdot \:5}$

$=y^{-3\cdot \:5}$

$=y^{-15}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

$=\frac{1}{y^{15}}$

4 0

3 years ago