All categories

3 years ago

Please help me to solve this question

Mathematics

1 answer:

densk [106]3 years ago

7 0

Answer:

(p+5)(p-2)

Step-by-step explanation:

We are looking for two numbers that multiply to -10 (the rightmost number) and sum to 3 (the middle number)

These are 5 and -2

So we write

(p ____)(p ____)

and fill in the blanks

(p+5)(p-2)

Check by FOILing:

p^2 -2p + 5p -10

And combine the two middle terms.

p^2 + 3p - 10

You might be interested in

Richard receives a 3.5% commission for selling cleaning supplies. What is his commission on sales of $122,591?

konstantin123 [22]

Answer:

We have to find 3.5% $122,591

So,

122591*3.5/100=4,290.685

So, the commission will be $4,290.685

5 0

3 years ago

Read 2 more answers

The area of a rectangle is 30 square root 3750 square inches, and the length is 3 square root 250 inches what is the width of th

Ann [662]

Answer:

$width=10\sqrt{15}\ in$

Step-by-step explanation:

-A rectangle's area is given by:

$A=lw\\\\l=length\\w=width$

Given A= $30\sqrt{3750}$ and $l=3\sqrt{250}$ , we substitute in the formula to solve for the width as follows:

$A=lw\\\\30\sqrt{3750}=3\sqrt{250}\times w\\\\w=\frac{30\sqrt{3750}}{3\sqrt{250}}\\\\w=10\frac{\sqrt{3750}}{\sqrt{250}}\\\\=10\times \sqrt{\frac{3750}{250}}\\\\=10\sqrt{15}\ in$

Hence , the rectangle's width is $10\sqrt{15}\ in$

6 0

3 years ago

I need help............

Alexxandr [17]

Problem 35

The bar graph is shown below

You simply draw various rectangles such that the heights represent the frequency of each animal type.

Eg: there are 20 elephants, so the "elephant" bar is 20 units tall.

You can make the bar graph by hand, or use spreadsheet software. I recommend going with software (if you can).

=======================================================

Problem 36

1 book = 20 mm thickness

5 books = 5*20 = 100 mm thickness

1 paper = 0.016 mm thickness

5 papers = 5*0.016 = 0.08 mm thickness

total thickness = 100+0.08 = 100.08 mm

<h3>Answer: 100.08 mm</h3>

=======================================================

Problem 37

121/11 = 11

121 = 11*11

If we say 11+11+11+...+11, and have 11 copies of these values added, then we get to a sum of 121

This is probably the easiest way to get the answer assuming repeated values are allowed.

You can stop here if your teacher allows you to use repeated values. If not, then move onto the next section.

-----------

If your teacher requires you to add 11 <u>different</u> numbers, then you can follow this procedure

Write out eleven copies of "11" in a sequence
Subtract 2 from the first "11" (to get 9) and add it to the last copy of "11" (to get 13)
Subtract 4 from the second "11" (to get 7) and add it to the second to last copy of "11" (to get 15)
Subtract 6 from the third "11" (to get 5) and add it to the third to last copy of "11" (to get 17)
Subtract 8 from the fourth "11" (to get 3) and add it to the fourth to last copy of "11" (to get 19)
Finally, subtract 10 from the fifth "11" (to get 1) and add it to the fifth to last copy of "11" (to get 21)

After carefully following those steps, you'll get this sequence (in the exact order shown):

{9, 7, 5, 3, 1, 11, 21, 19, 17, 15, 13}

There are three properties to notice of this sequence

It's composed of two decreasing arithmetic sequences {9, 7, 5, 3, 1} and {21, 19, 17, 15, 13}
The 11 in the middle hasn't been changed from the original sequence of nothing but "11"s.
You should find that the terms of this new sequence add to 121.

That sequence we got sorts to {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21}

and we can say 1+3+5+7+9+11+13+15+17+19+21 = 121

<h3>Answer: 1+3+5+7+9+11+13+15+17+19+21 = 121</h3>