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3 years ago

What is 5 times 5 plus 67 divided by 0 plus 56?

2 answers:

8 0

(5 x 5 + 67) / (0 + 56)
<span>(5 x 5 + 67) / 56 (The zero is irrelevant.)
</span><span>(25 + 67) / 56 (5x5 = 25)
</span>92/56 (25+67=92)
1.64 (92/56 rounded to 2 decimal places.)

Note that this answer assumes the format shown in the first line, specifically regarding the parenthesis. If there are some in a different format (say instead 5 x (5 + 67)) then the answer will be different.

kiruha [24]3 years ago

5 0

Answer:56

Step-by-step explanation: A faster way to do it is too not do 5x5 plus 67 because after that you divided by 0. If you divide by 0 you get nothing. So next is 0 plus 56. 0 means nothing so just let 56 be itself.

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What is the equation of the line that passes through the point (5, 4) and is perpendicular to the line whose equation is 2x + y

konstantin123 [22]

So a line that is perpendicular is a line that intercepts other line at a 90 degree angle.

Slope of perpendicular lines is the opposite sign and the reciprocal of the other.

So if your slope was 2 the perpendicular line would have a slope of -1/2.

So the equation, 2x + y = 3

Can be rewritten as

y= -2x+3

Perpendicular slope: 1/2

Since it intercepts/ passes the coordinates (5,4) it means that those values are included in the function input x and output y.

y=mx+b

4=1/2(5)+b

Solve for b (y intercept)

4-2.5=b

1.5=b

So the equation of the perpendicular is

y=1/2x+3/2 or y=.5x+1.5

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3 years ago

Why are the solutions to the proportion‘s 50 over X equals 10/20 and 10/50 equals 20 over X the same?

Ahat [919]

Answer

both get the same equation 10x=1000 which simplifies to x=100. HOPE THIS HELPED!!!

Step-by-step explanation:

Do I really need to?

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3 years ago

Growth Models 19515. In 1968, the U.S. minimum wage was $1.60 per hour. In 1976, the minimum wagewas $2.30 per hour. Assume the

VikaD [51]

In general, the exponential growth function is given by the formula below

$f(x)=a(1+r)^x$

Where a and r are constants, and x is the number of time intervals.

In our case, n=0 for 1960; therefore, 1968 is n=8,

$\begin{gathered} f(8)=a(1+r)^8 \\ \text{and} \\ f(8)=1.6 \\ \Rightarrow1.6=a(1+r)^8 \end{gathered}$

And 1976 is n=16

$\begin{gathered} f(16)=a(1+r)^{16} \\ \text{and} \\ f(16)=2.3 \\ \Rightarrow2.3=a(1+r)^{16} \end{gathered}$

Solve the two equations simultaneously, as shown below

$\begin{gathered} \frac{1.6}{(1+r)^8}=a \\ \Rightarrow2.3=\frac{1.6}{(1+r)^8}(1+r)^{16} \\ \Rightarrow2.3=1.6(1+r)^8 \\ \Rightarrow\frac{2.3}{1.6}=(1+r)^8 \\ \Rightarrow(\frac{2.3}{1.6})^{\frac{1}{8}}=(1+r)^{}^{} \\ \Rightarrow r=(\frac{2.3}{1.6})^{\frac{1}{8}}-1 \\ \Rightarrow r=0.0464078 \end{gathered}$

Solving for a,

$\begin{gathered} r=0.0464078 \\ \Rightarrow a=\frac{1.6}{(1+0.0464078)^8}=1.113043\ldots \end{gathered}$

a) Thus, the equation is

$\Rightarrow f(n)=1.113043\ldots(1+0.0464078\ldots)^n$

b) 1960 is n=0; thus,

$f(0)=1.113043\ldots(1+0.0464078\ldots)^0=1.113043\ldots$

The answer to part b) is $1.113043... per hour

c)1996 is n=36

$\begin{gathered} f(36)=1.113043\ldots(1+0.0464078\ldots)^{36} \\ \Rightarrow f(36)=5.6983\ldots \end{gathered}$

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1 year ago

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