Write ACUTE, RIGHT, OBTUSE, STRAIGHT or REFLEX for each of the following: ( 5 marks)

Factor completely 15x^5 + 12x^4-24x^3-3x^2

diamong [38]

Answer:

3x^2 (5x^3 + 4x^2 - 8x - 1)

Step-by-step explanation:

The given expression is 15x^5 + 12x^4-24x^3-3x^2

Here we have to find the GCF of all the above terms.

The GCF is 3x^2, let's take out 3x^2 and write the remaining terms in the parenthesis.

15x^5 + 12x^4-24x^3-3x^2

=3x^2 (5x^3 + 4x^2 - 8x - 1)

Therefore, the factors are 3x^2 and (5x^3 + 4x^2 -8x -1).

15x^5 + 12x^4-24x^3-3x^2 = 3x^2 (5x^3 + 4x^2 -8x -1)

Thank you.

7 0

2 years ago

Paige measured 4/9 cup of onions, 3/4 cup of celery, and 2/7 cup of carrots. How many cups of vegetables did Paige measure out f

lozanna [386]

1whole 121/252
sorry if it's wrong

3 0

2 years ago

Read 2 more answers

Moby sold half his toy car collection, and then bought 12 more cars the next day. He now has 48 toy cars. How many cars did he h

castortr0y [4]

Answer: 72 cars.

Step-by-step explanation:

Let be "x" the number of toys that Moby had to start with.

According to the information given in the exercise, Moby sold half his toy car collection. This can be represented with the following expression:

$\frac{1}{2}x$

Then he bought 12 more cars, giving a total amount of 48 toy cars.

Therefore, the equation that represents this situation is the equation shown below:

$x-\frac{1}{2}x+12=48$

Now you must solve for "x" in order to find its value.

You get that this is:

$\frac{1}{2}x+12=48\\\\\frac{1}{2}x=48-12\\\\\frac{1}{2}x=36\\\\x=(36)(2)\\\\x=72$

7 0

3 years ago

What is x2 – 6 = 16x + 30 rewritten in standard form, then factored?

snow_lady [41]

The answer is
x=18,-2

6 0

2 years ago

Sunspots have been observed for many centuries. Records of sunspots from ancient Persian and Chinese astronomers go back thousan

Lunna [17]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ $\leq$ 41

Alternate Hypothesis, $H_A$ : $\mu$ > 41

(b) The value of z test statistics is 1.08.

(c) We conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

Step-by-step explanation:

We are given that in a random sample of 40 such periods from Spanish colonial times, the sample mean is x¯ = 47.0. Previous studies of sunspot activity during this period indicate that σ = 35.

It is thought that for thousands of years, the mean number of sunspots per 4-week period was about µ = 41.

Let $\mu$ = mean sunspot activity during the Spanish colonial period.

(a) Null Hypothesis, $H_0$ : $\mu$ $\leq$ 41 {means that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41}

Alternate Hypothesis, $H_A$ : $\mu$ > 41 {means that the mean sunspot activity during the Spanish colonial period was higher than 41}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean = 47

σ = population standard deviation = 35

n = sample of periods from Spanish colonial times = 40

So, the test statistics = $\frac{47-41}{\frac{35}{\sqrt{40} } }$

= 1.08

(b) The value of z test statistics is 1.08.

(c) Now, the P-value of the test statistics is given by;

P-value = P(Z > 1.08) = 1 - P(Z < 1.08)

= 1 - 0.8599 = 0.1401

Since, the P-value of the test statistics is higher than the level of significance as 0.1401 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

8 0

2 years ago