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Julli [10]
3 years ago
12

When this polynomial is divided by (m+1), the remainder is 0. What is the value of the polynomial’s constant term?

Mathematics
2 answers:
klasskru [66]3 years ago
6 0

Answer:

-2m^2+3m-4

Step-by-step explanation:

\frac{-2m^3+m^2-m-4}{m+1}

avanturin [10]3 years ago
6 0

Answer: it would be A. -4

Step-by-step explanation:

Got it right on the test

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Write the equation of the graph
olchik [2.2K]

Answer:

y = 2.cos(x) - 2

Step-by-step explanation:

1. Shifting cos(x) right or left, top or down, as the graph in the question has the same illustration of  cos(x). Consider

y = a.cos(x) + b

2. find a and b by using two sample points from the graph: (0, 0) and (\pi, -4)

0 = a.cos(0) + b  => a + b = 0\\ -4 = a.cos(\pi) + b => b-a=-4\\\left \{ {{2b=-4} \atop {2a=4}} \right.\\a = 2, b=-2\\

5 0
3 years ago
Fraction simplifier de <img src="https://tex.z-dn.net/?f=%5Cfrac%7B15%7D%7B35%7D" id="TexFormula1" title="\frac{15}{35}" alt="\f
Pavel [41]
You have to divide them with 5
3/7
3 0
3 years ago
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let <em>X</em> = number of students who read above the eighth grade level.

(a)

A sample of <em>n</em> = 269 students are selected. Of these 269 students, <em>X</em> = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

\hat p=\frac{X}{n}=\frac{224}{269}=0.8327

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

1-\hat p=1-0.8327

        =0.1673

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

<em>n</em> = 709

<em>X</em> = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

\hat q=1-\hat p

  =1-\frac{X}{n}

  =1-\frac{546}{709}

  =0.2299\\\approx 0.229

The critical value of <em>z</em> for 95% confidence interval is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the 95% confidence interval for the population proportion as follows:

CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

     =0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

5 0
3 years ago
PLEASE HELPPP, its timed. NO BOTS please i want serious answers.
Rzqust [24]
CDAB

4.2 (convert to frac on ur own)

12.5=x

CB= 14.5

AB= 8.82
6 0
3 years ago
Which postulate or theorem proves that these two triangles are congruent?
Alla [95]

AAS congruence theorem.


We know that <H is congruent to <F and <GJH is congruent to <JGF.

We also know that JG is congruent to JG, which gives us a side and two angles, so AAS would prove them congruent.

4 0
3 years ago
Read 2 more answers
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