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2 years ago

What is 7.4 + ? = 1.

Mathematics

1 answer:

anastassius [24]2 years ago

5 0

?= -6.4. The number, negative 6.4 would go where the question mark is.

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I need help with this i need to find the grey area

IrinaVladis [17]

Answer:

y < 4x - 2

Step-by-step explanation:

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2 years ago

Read 2 more answers

Select all of the solutions to the system graphed below.

ra1l [238]

0,-2

4,4

2,0

3,5

Step-by-step explanation:

that was my answer

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3 years ago

An employee compiled sales data for a company once each month. The scatter plot below shows the sales (in multiples of $1000) fo

svp [43]

Answer:

940

Step-by-step explanation:

The scatter plot below shows the sales (in multiples of $1000) for the company over time (in months).

Also the sales can be modeled by the help of a linear function as:

y = 0.94x + 12.5.

Now we know that the company's sales increase per month is the slope of the linear function by which this situation is modeled.

We know that for any linear function of the type:

y=mx+c

'm' represents the slope and 'c' represents the y-intercept of the line.

Hence, by looking at the equation we get:

m=0.94

but as the sales are multiplied by 1000.

Hence,

0.94×1000=$ 940.

Hence, the company's sales increase per month is:

$ 940

5 0

2 years ago

Evaluate the function f(x) at the given numbers (correct to six decimal places).

statuscvo [17]

Answer:

The values of given function are shown in the below table.

Step-by-step explanation:

The given function is

$f(x)=\frac{x^2-5x}{x^2-x-20}$

Simplify the given function.

$f(x)=\frac{x(x-5)}{x^2-5x+4x-20}$

$f(x)=\frac{x(x-5)}{x(x-5)+4(x-5)}$

$f(x)=\frac{x(x-5)}{(x+4)(x-5)}$

Cancel out the common factor.

$f(x)=\frac{x}{x+4}$

Substitute x=5.5 in the above equation.

$f(5.5)=\frac{5.5}{5.5+4}$

$f(5.5)=\frac{5.5}{9.5}$

$f(5.5)=0.57894736842$

$f(5.5)\approx 0.578947$

Similarly find the value for all values of x.

The values of given function are shown in the below table.

8 0

3 years ago

Two vertical poles of length 10 and 12 feet, respectively, stand 17 feet apart. A cable reaches from the top of one pole to the

Marat540 [252]

Answer:

$L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433} ft$

Step-by-step explanation:

let length of calble = L

distance from 10-foot pole = x ft

Moving from the top of the 10-foot pole to point x, we have

From the diagram that

$x^{2} +10^{2}=L_{1} ^{2} \\x^{2} + 100 = L_{1} ^{2}$

Take the square root of both sides

$\sqrt{x^{2}+100}=\sqrt{L_{1} ^{2}}\\ \sqrt{x^{2}+100}= L_{1}$

Moving from point x to the top of the 12-foot pole, we have

From the diagram that

$(17-x)^{2}+12^{2}=L_{2} ^{2} \\289-34x+x^{2} +144=L_{2} ^{2}\\x^{2} -34x+433=L_{2} ^{2}$

Take the square root of both sides

$\sqrt{x^{2}-34x+433}=\sqrt{L_{2} ^{2}} \\\sqrt{x^{2}-34x+433}=L_{2}$

Amount of cable used, L = L1 + L2

$L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433}$

3 0

3 years ago