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3 years ago

Please help and explain!

Mathematics

1 answer:

Sphinxa [80]3 years ago

5 0

I think I'm not sure sorry if it's wrong

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-7+2 Chises: 6 5 -5 9

Allushta [10]

The answer is -5 isjsjshsj

3 0

3 years ago

Are multiples of 3 are always multiples of 6? True or False.

MissTica

Answer:

False

Step-by-step explanation:

9 is a multiple of 3 but not 6

7 0

3 years ago

Read 2 more answers

Can u help me ASAP. i need to know how to do it step by step

12345 [234]

Answer:

19

Step-by-step explanation:

$f(x) =4x^3-8x^2+ax+b$ has a factor $2x-1$ and

when divided by $x+2$ , remainder is 20.

To find:Remainder when divided by (x-1) ?

Solution:

$2x-1$ is a factor

$2x - 1 = 0 \Rightarrow x = \frac{1}{2}$ when we put this value of x to the function, it will become 0.

i.e.

$\Rightarrow f(\dfrac{1}{2}) =0 =4\times (\dfrac{1}2)^3-8(\dfrac{1}2)^2+\dfrac{a}{2}+b=0\\\Rightarrow \dfrac{1}{2}-2+\dfrac{a}{2}+b=0\\\Rightarrow 1-4+a+2b=0\\\Rightarrow a +2b=3 ......(1)$

Remainder is 20 when f(x) is divided by $x+2$

i.e.

$f(-2) =20$

$\Rightarrow f(-2) =20 =4\times (-2)^3-8(-2)^2-2a+b=20\\\Rightarrow -32-32-2a+b=20\\\Rightarrow -2a+b=84 ...... (2)$

Solving (1) and (2), Multiply equation (1) by 2 and adding to (2):

$5b=6+84\\\Rightarrow b = \dfrac{90}{5} = \bold{18}$

By equation (1):

$a+2(18) = 3\\\Rightarrow a = -33$

So, the equation becomes:

$f(x) =4x^3-8x^2-33x+18$

$\Rightarrow f(1) = 4(1) -8 (1) -33(1) +18 = \bold{19}$

So, when divided by (x-1), remainder will be 19.

4 0

3 years ago

Juma earns $12.50 for each newspaper subscription he sells. He also earns $50 base salary each week. If he wants to earn $400 ne

Mandarinka [93]

He would need to sell 28 because
12.50*28=350 and 350+50=400

4 0

3 years ago

A , B , C , D ?????????

3241004551 [841]

Answer:

Step-by-step explanation:

the question is asking for the difference between the two values, therfore, you should subtract the smaller value from the larger value

while A is technically the same equation, when finding a value, you want all the constants to be on 1 side of the equation while the variable is on the other side

5 0

2 years ago