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3 years ago

Please help explain on how to do it as I am having trouble solving it.

Mathematics

1 answer:

aleksley [76]3 years ago

5 0

Answer:

the value if x is 15

Step-by-step explanation:

Use Pythagoras theorem

x² = (x - 3)² + 9²

x² = x² -3x -3x + 9 + 81

x² = x² -6x + 90

x² - x² + 6x = 90

6x = 90

divide through by 6

x = 90/6

x = 15

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Draw the model to show 5.5/5

babunello [35]

This would be easy if you made more sense out of the question.

4 0

3 years ago

54 students failed a math test. This is 27% of the total number of students. How many students passed the test?

nikdorinn [45]

Answer:

7/3 or 2 1/3

Step-by-step explanation:

(2x3) + (1+1)

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1x3

Hope this helps

3 0

3 years ago

3. The profit for a company is given by the equation y = -0.0002x 2 + 140x – 250,000 where x is the number of units sold and y i

nirvana33 [79]

9514 1404 393

Answer:

350,000

Step-by-step explanation:

The vertex of the quadratic ax²+bx+c is located at x=-b/(2a). For the given profit function, the number of units (x) that will produce maximum profit is ...

x = -(140)/(2(-0.0002)) = 1,400,000/4

x = 350,000

350,000 units should be sold to maximize profit.

3 0

3 years ago

prove that if f is integrable on [a,b] and c is an element of [a,b], then changing the value of f at c does not change the fact

Neko [114]

Answer with Step-by-step explanation:

We are given that if f is integrable on [a,b].

c is an element which lie in the interval [a,b]

We have to prove that when we change the value of f at c then the value of f does not change on interval [a,b].

We know that limit property of an integral

$\int_{a}^{b}f dt=\int_{a}^{c}fdt+\int_{c}^{b} fdt$

$\int_{a}^{b} fdt=f(b)-f(a)$ ....(Equation I)

Using above property of integral then we get

$\int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b} fdt$ ......(Equation II)

Substitute equation I and equation II are equal

Then we get

$\int_{a}^{b}fdt= f(c)-f(a)+{f(b)-f(c)}$

$\int_{a}^{b}fdt=f(c)-f(a)+f(b)-f(c)=f(b)-f(a)$

$\int_{a}^{c}fdt+\int_{c}^{b}fdt=f(b)-f(a)$

Therefore, $\int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b}fdt$ .

Hence, the value of function does not change after changing the value of function at c.

6 0

3 years ago

Identify the days when the temperature was 76°F or more.

N76 [4]

Answer: Tuesday, Thursday, and Saturday

Step-by-step explanation:

Pretty self-explanatory.

3 0

2 years ago

Read 2 more answers

Please help explain on how to do it as I am having trouble solving it.​

Please help explain on how to do it as I am having trouble solving it.