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spayn [35]
3 years ago
8

What is the answer to the question below, I have been working on it for a while. Plz help!

Mathematics
1 answer:
Vsevolod [243]3 years ago
3 0

Answer:

10m / 7

Step-by-step explanation:

times = multiplication of 10 and variable m

quotient of = division of 7

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What is y squared + 2 y - 35 / y + 7​
lutik1710 [3]

Answer:

y - 5

Step-by-step explanation:

Given

\frac{y^2+2y-35}{y+7}

Factorise the numerator

y² + 2y - 35 = (y + 7)(y - 5), thus fraction can be expressed as

\frac{(y+7)(y-5)}{y+7}

Cancel the factor (y + 7) on the numerator/ denominator, leaving

y - 5

5 0
3 years ago
Read 2 more answers
Determine whether the equation represents a direct variation. If it does, find the constant of variation. 7x = −6y
s2008m [1.1K]
The constant variation -7/6
5 0
3 years ago
Read 2 more answers
If a dozen exercise books cost 144, what will 14 of the same books cost
Aleksandr-060686 [28]
168


So you take the 144 divide 12(the dozen) so we could find how much 1 cost, it was 12. You added 2 more books so we add(12x2)+(144). That equals to 168

Hope this helps!
8 0
2 years ago
Question
Kay [80]

Answer:

f(x)=2(x-2)(x^2+64)

Step-by-step explanation:

A standard polynomial in factored form is given by:

f(x)=a(x-p)(x-q)...

Where <em>p</em> and <em>q</em> are the zeros.

We want to find a third-degree polynomial with zeros <em>x</em> = 2 and <em>x </em>= -8i and equals 320 when <em>x </em>= 4.

First, by the Complex Root Theorem, if <em>x</em> = -8i is a root, then <em>x </em>= 8i must also be a root.

Therefore, we acquire:

f(x)=a(x-(2))(x-(-8i))(x-(8i))

Simplify:

f(x)=a(x-2)(x+8i)(x-8i)

Expand the second and third factors:

=(x+8i)x+(x+8i)(-8i)\\\\=(x^2+8ix)+(-8ix-64i^2)\\\\=(x^2)+(8ix-8ix)+(-64i^2)\\\\=x^2-64(-1)\\\\ =x^2+64

Hence, our function is now:

f(x)=a(x-2)(x^2+64)

It equals 320 when <em>x</em> = 4. Therefore:

320=a(4-2)(4^2+64)

Solve for <em>a</em>. Evaluate:

320=(2)(80)a

So:

320=160a\Rightarrow a=2

Our third-degree polynomial equation is:

f(x)=2(x-2)(x^2+64)

7 0
3 years ago
There are 11 students on the tennis team. The coach selects 3 of them to go to atennis clinic. In how many ways can he choose 3
Komok [63]

Answer:

165 ways

Step-by-step explanation:

Selection deals with combination

There are a total of 11 from which 3 are to be selected

        11C3 = 11!/3!(11-3)!

                 = 11!/(3!x8!)

                 =(11x10x9x8!)/(3x2x8!)

                 =11x10x9/6

                 =11x5x3 = 165 ways

                 

8 0
3 years ago
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