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Maurinko [17]
3 years ago
8

Can someone help me out

Mathematics
1 answer:
tankabanditka [31]3 years ago
3 0

Answer:

I think that your answer is 1 .

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What is the value of X? Enter your answer in the box
dmitriy555 [2]

Answer:

\large\boxed{x=26}

Step-by-step explanation:

\text{The angle}\ x^o\ \text{and}\ 64^o\ \text{together from both sidesr form a right angle.}\\\text{Therefore we have the equation:}\\\\x^o+64^o=90^o\qquad\text{subtract}\ 64^o\ \text{from both sides}\\\\x^o+64^o=90^o-64^o\\\\x^o=26^o

4 0
3 years ago
Read 2 more answers
Can someone help me out?
Anarel [89]

Answer:

So the value of s  = 7 cm

Step-by-step explanation:

We know that, Area of a triangle = 1/2 * b * h

So, base = 24 cm

height = s

Area = 1/2 * b * h

=>84 = 1/2 * 24 * h

=>84 = 12 * h

=>84 / 12 = h

=> 7 = h (in cm)

Hope this helps, please mark me as brainliest!! Thank you!

6 0
4 years ago
the length of a rectangle is two feet greater then twice its width if the primeter is 25 find the width
Mademuasel [1]

Answer:

Step-by-step explanation:

Perimeter is simply length + width. They gave us that the length is 2 feet plus 2 times the width:

25=(2w+2)+W

Subtract 2 from both sides to get 23 = 2w + w

Combine like terms to get 23 = 3w

Divide by 3 to get w=7 \frac{2}3} feet

7 0
3 years ago
If f(1) = 0, what are all the roots of the function f(x)=x^3+3x^2-x-3? Use the Remainder Theorem.
Sophie [7]
There's no if about it, 

f(x)=x^3+3x^2-x-3


has a zero f(1)=0 so x-1 is a factor.   That's the special case of the Remainder Theorem; since f(1)=0 we'll get a remainder of zero when we divide f(x) by x-1.

At this point we can just divide or we can try more little numbers in the function.  It doesn't take too long to discover f(-1)=0 too, so  x+1 is a factor too by the remainder theorem.  I can find the third zero as well; but let's say that's out of range for most folks.

So far we have 

x^3+3x^2-x-3 = (x-1)(x+1)(x-r)

where r is the zero we haven't guessed yet.  Again we could divide f(x) by (x-1)(x+1)=x^2-1 but just looking at the constant term we must have

-3 = -1 (1)(-r) = r

so

x^3+3x^2-x-3 = (x-1)(x+1)(x+3)

We check f(-3)=(-3)^3+3(-3)^2 -(-3)-3 = 0 \quad\checkmark

We usually talk about the zeros of a function and the roots of an equation; here we have a function f(x) whose zeros are

x=1, x=-1, x=-3

8 0
3 years ago
Read 2 more answers
Choose the product. -5p 3(4p 2 + 3p - 1) -20p 5 - 15p 4 + 5p 3 -20p 6 - 15p 4 - 5p 3 -20p 5 + 3p - 1 20p 5 + 15p 4 - 5p 3
Likurg_2 [28]
<span>-5p^3(4p^2 + 3p - 1) = -20p^5 - 15p^4 + 5p^3

answer is

A. first one
  </span>-20p^5 - 15p^4 + 5p^3
5 0
3 years ago
Read 2 more answers
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