If 40% of a number is 112 and 95% of the same number is 266, find 55% of that number.

Use polynomial division to completely factor y=x^3+3x^2-13x-15 by x+5

Salsk061 [2.6K]

Answer:

(x - 3)(x + 1)(x + 5)

Step-by-step explanation:

I'd use synthetic division instead. If we were to find the roots of the given polynomial, we could from them write the factors as well.

The divisor x + 5 corresponds to root x = -5. Setting up synthetic div.,

-5 ) 1 3 -13 -15

-5 10 +15

-----------------------------

1 -2 -3 0

Since the remainder is 0, we know that -5 is a root and (x + 5) is a factor. Moreover, we know that the coefficients of the quotient are 1, -2 and -3.

1x² - 2x - 3 can be factored: the factors are (x - 3) and (x + 1).

So the end result for this problem is (x - 3)(x + 1)(x + 5).

7 0

3 years ago

-11y = 6(2+1) - 13y

iris [78.8K]

Answer:

12

Step-by-step explanation:

Let's put the equations in standard form. For the first equation, we have:

−11y=6(z+1)-13y

2y−6z=6

y−3z=3

The second equation is:

4y−24=c(z−1)

4y−cz=24−c

If we multiply the first equation by 4, we get:

4y-12z=12

Comparing the two equations, we see that if c=12, both equations will be the same and there will be infinitely many solutions.

The correct value of c is 12.

7 0

3 years ago

Generate the nest three terms of each arithmetic sequence shown below.

o-na [289]

Answer:

A)2,6,10

B)2,-6,-18

C)-1,-3,-5

Step-by-step explanation:

A)a1=-2 and d=4

We know that the arithmetic sequence formula is

$a_{n}=a_{1}+(n-1)d$

Now

$a_{2}=a_{1}+(2-1)d$

$a_{2}=a_{1}+(1)d$

Substituting the given value we get

$a_{2}= -2+(1)4$

$a_{2}= -2+4$

$a_{2}= 2$

------------------------------------------

Similarly

$a_{3}=a_{1}+(3-1)d$

$a_{3}=a_{1}+(2)d$

Substituting the given value we get

$a_{3}= -2+(2)4$

$a_{3}= -2+8$

$a_{3}= 6$

-------------------------------------------------

$a_{4}=a_{1}+(4-1)d$

$a_{4}=a_{1}+(3)d$

Substituting the given value we get

$a_{4}= -2+(3)4$

$a_{4}= -2+16$

$a_{4}= 10$

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B $a_n=a_{(n-1)}-8$ with $a_1=10$

$a_2=a_{(2-1)}-8$

$a_2=a_{1}-8$

Substituting the given value

$a_2= 10-8$

$a_2=2$

---------------------------------------------------------------------

$a_3=a_{(3-1)}-8$

$a_3=a_{2}-8$

Substituting the value

$a_3=2-8$

$a_3= -6$

---------------------------------------------------------------------

$a_4=a_{(4-1)}-8$

$a_4=a_{3}-8$

Substituting the value

$a_4= -6-8$

$a_4= -14$

-------------------------------------------------------------------------------------------

C) $a_1=3, a_2=1$

Here the difference is -2

the arithmetic sequence formula is

$a_{n}=a_{1}+(n-1)d$

Now

$a_{3}=a_{1}+(3-1)d$

$a_{3}=a_{1}+(2)d$

Substituting the value we get

$a_{3}= 3+(2)-2$

$a_{3}= 3-4$

$a_{3}= -1$

------------------------------------------------------------------------------

$a_{4}=a_{1}+(4-1)d$

$a_{4}=a_{1}+(3)d$

Substituting the value we get

$a_{4}= 3+(3)-2$

$a_{4}= 3-6$

$a_{4}= -3$

----------------------------------------------------------------------------------

$a_{5}=a_{1}+(5-1)d$

$a_{5}=a_{1}+(4)d$

Substituting the value we get

$a_{5}= 3+(4)-2$

$a_{5}= 3-8$

$a_{5}= -5$

4 0

3 years ago

The number 2.31 x 10 to the power of x is 2,200 times greater than 1.05 x 10 to the power of y.

Goshia [24]

Answer:

x - y = 3

any values of x& y where x is three greater than y

I.E. x = 8 y = 5

Step-by-step explanation:

8 0

3 years ago

What fraction of the students in the class named baseball their favorite sport

mojhsa [17]

now, let's take a peek at the denominators, we have 3 and 8 and 12, we can get an LCD of 24 from that.

Let's multiply both sides by the LCD of 24, to do away with the denominators.

so, let's recall that a whole is "1", namely 500/500 = 1 = whole, or 5/5 = 1 = whole or 24/24 = 1 = whole. So the whole class will yield a fraction of 1/1 or just 1.

$\bf ~\hspace{7em}\stackrel{\textit{basketball}}{\cfrac{1}{3}}+\stackrel{\textit{soccer}}{\cfrac{1}{8}}+\stackrel{\textit{football}}{\cfrac{5}{12}}+\stackrel{\textit{baseball}}{x}~=~\stackrel{\textit{whole}}{1} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{24}}{24\left(\cfrac{1}{3}+\cfrac{1}{8}+\cfrac{5}{12}+x \right)=24(1)}\implies (8)1+(3)1+(2)5+(24)x=24 \\\\\\ 8+3+10+24x=24\implies 21+24x=24\implies 24x=3 \\\\\\ x=\cfrac{3}{24}\implies x=\cfrac{1}{8}$

4 0

3 years ago