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3 years ago

Cuales son los lados que faltan? Puede alguien ayudarme!??

Mathematics

1 answer:

Archy [21]3 years ago

3 0

Answer:

Yo elegiría la respuesta ya seleccionada

Step-by-step explanation:

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Suppose you travel at 50 mph for 201 miles. How many hours will it take you to reach your destination? A.5.3 hours B.4.02 hours

Kamila [148]

201 miles ÷ 50 mph = 4.02 hours (B)

3 0

4 years ago

Read 2 more answers

Nick starts with 20 milligrams of a radioactive substance. The amount of the substance decreases by 14 each week for a number of

zlopas [31]

Answer:

Eric's expression is :

$20(\frac{1}{4})^w$

and Andrea's is :

$(1-0.5)^w$

In Eric's expression, 20 represents the initial amount of substance with which he has started the experiment. $\thinspace \frac{1}{4}\thinspace$ is the amount of substance left after each time period (in this case, each week).The variable w in this case represents the number of weeks.

Andrea's expression can be written as :

$1\cdot (1-0.5)^w$

The one outside of parentheses represents the initial amount of the substance. The one inside of parentheses represents 100% of the original amount of the substance. 0.5 represents the 50% of the substance that is lost each time period. The variable w in this case represents the number of weeks.

8 0

3 years ago

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Greg bought 2 boxes of balloons. He used half of them to decorate his yard. he used 40 to decorate his porch. and used the rest

Anuta_ua [19.1K]

20 I'm guessing? Hope this helps!

6 0

4 years ago

Consider the equation below.

Korvikt [17]

Answer:

Equation in square form:

$y=3(x+5)^2-4$

Extreme value:

$(h,k)=(-5,-4)$

Step-by-step explanation:

We are given

$y=3x^2+30x+71$

we can complete square

$y=3(x^2+10x)+71$

we can use formula

$a^2+2ab+b^2=(a+b)^2$

$y=3(x^2+2\times x\times 5)+71$

now, we can add and subtract 5^2

$y=3(x^2+2\times x\times 5+5^2-5^2)+71$

$y=3(x^2+2\times x\times 5+5^2)-3\times 5^2+71$

$y=3(x+5)^2-75+71$

So, we get equation as

$y=3(x+5)^2-4$

Extreme values:

we know that this parabola

and vertex of parabola always at extreme values

so, we can compare it with

$y=a(x-h)^2+k$

where

vertex=(h,k)

now, we can compare and find h and k

$y=3(x+5)^2-4$

we get

h=-5

k=-4

so, extreme value of this equation is

$(h,k)=(-5,-4)$

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4 years ago

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Solve an equation for one variable

AysviL [449]

To solve an equation for one variable, we will utilize PEMDAS ( parentheses, exponents, multiplication/division, add/subtract) but backwards. So we will first add/subtract any numbers to the other side to get the variable by itself. We will then divide/multiply any numbers to get the variable by itself.

7 0

2 years ago