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3 years ago

Two years of local Internet service cost $685, including the installation fee of $85. What is the monthly fee?

Mathematics

2 answers:

AnnZ [28]3 years ago

8 0

685-85=600
1 year =12 months
12 multiply by 2 =24
600 divided by 24=25
answer=25

guapka [62]3 years ago

7 0

The answer is 770 dollars!
Hope this helped!!
-EmmyLou2020

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Amanda [17]

Answer:

$40\ bottle\ caps$

Step-by-step explanation:

we know that

The scale is $\frac{2.4}{1}\ \frac{cm}{ft}$

That means

2.4 cm in the model represent 1 foot in the real

In this problem

The restaurant is 20 feet tall

Find the height of the model

Multiply the actual height of the restaurant by the scale 2.4

$20\ ft=20(2.4)=48\ cm$

A bottle cap is 1.2 cm tall

To find out the number of bottle caps divide the height of the model by 1.2

$\frac{48}{1.2}= 40\ bottle\ caps$

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3 years ago

Kerry has a new phone plan. It works by charging $20 per month and then $0.10 for each minute over 200 minutes.

REY [17]

Answer:

(40-20)/.10

Step-by-step explanation:

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2 years ago

Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

Naily [24]

Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

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$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

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3 years ago

Enrique has to pay 80% of a $200 phone bill. Explain how to use equivalent ratios to find 80% of $200.

olga2289 [7]

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3 years ago

Read 2 more answers

Altitudes AA1 and BB1 are drawn in acute △ABC. Prove that A1C·BC=B1C·AC

Sophie [7]

Answer:

See the attached figure which represents the problem.

As shown, AA₁ and BB₁ are the altitudes in acute △ABC.

△AA₁C is a right triangle at A₁

So, Cos x = adjacent/hypotenuse = A₁C/AC ⇒(1)

△BB₁C is a right triangle at B₁

So, Cos x = adjacent/hypotenuse = B₁C/BC ⇒(2)

From (1) and (2)

∴ A₁C/AC = B₁C/BC

using scissors method

∴ A₁C · BC = B₁C · AC

7 0

3 years ago