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3 years ago

Equation word problem

Mathematics

2 answers:

wolverine [178]3 years ago

3 0

Answer:

190

Step-by-step explanation:

40 + 70 + 80 = 190

Sedaia [141]3 years ago

3 0

I think it’s 190 I’m not sure tho but you can double check

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2/3y + 15 = 9 What are the steps for this question

xenn [34]

Answer:

see below

Step-by-step explanation:

2/3y + 15 = 9

Subtract 15 from each side

2/3y + 15-15 = 9-15

2/3y = -6

Multiply each side by 3/2 to isolate y

3/2 * 2/3y = -6 *3/2

y = -9

4 0

3 years ago

Read 2 more answers

Change 0.7 to a common fraction

erma4kov [3.2K]

Answer:

7/10

Step-by-step explanation:

In 0.7 we will change the decimal to fraction. First we will write the decimal without the decimal point as the numerator. Now in the denominator, write 1 followed by one zeros as there are 1 digit in the decimal part of the decimal number. Therefore, we observe that 0.7 (decimal) is converted to 7/10 (fraction).

3 0

3 years ago

Last year, a widget factory produced one million,twele thousand, sixty widgets. What is the nunber in standard form

aivan3 [116]

1,012,060 this is it in standard form

6 0

3 years ago

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Suppose you have developed a scale that indicates the brightness of sunlight. Each category in the table is 4 times brighter tha

raketka [301]

According to the developed scale, a radiant day is 16 times brighter than a dim day.

We assume the brightness of a dim day to be x.

According to the developed scale, the brightness of an illuminated day will be 4 times that of a dim day.

Thus, the brightness of an illuminated day = 4*the brightness of a dim day = 4x.

According to the developed scale, the brightness of a radiant day will be 4 times that of an illuminated day.

Thus, the brightness of a radiant day = 4*the brightness of an illuminated day = 4*4x = 16x.

Now, the ratio of the brightness of a radiant day to the brightness of a dim day = 16x:x = 16x/x = 16:1.

Thus, according to the developed scale, a radiant day is 16 times brighter than a dim day.

Learn more about the developed scale at

brainly.com/question/4970963

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The question provided is incomplete. The complete question is:

"Suppose you have developed a scale that indicates the brightness of sunlight. Each category in the table is 4 times brighter than the next lower category. For example, a day that is dazzling is 4 times brighter than a day that is radiant. How many times brighter is a radiant day than a dim day?

Dim=2

Illuminated=3

Radiant=4

Dazzling=5"

8 0

1 year ago

If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.

Pie

Given that $f^{(n)}(0)=(n+1)!$ , we have for $f(x)$ the Taylor series expansion about 0 as

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n$

Replace $n+1$ with $n$ , so that the series is equivalent to

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}$

and notice that

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}$

Recall that for $|x|$ , we have

$\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}$

which means

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}$
$\implies f(x)=\dfrac1{(1-x)^2}$

5 0

3 years ago