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3 years ago

6x -2x3 + 10x+2x3+7x4 simply

2 answers:

4 0

$(((6x-(2*(x3)))+10x)+(2*(x3)))+7x4 \\ \\ Then \ we \ simplify \ it \ down \ more. \\ \\ (((6x-(2*(x^3)))+10x)+2x^3)+7x^4 \\ \\ then \ we \ would \ then \ pull \ out \ the \ like \ factors . . . . \\ \\ 7x^4 + 16x = x * (7x^3 + 16) \\ \\$

Your final answer: $\boxed{\bf{ x *(7x^3 + 16)}}$

VMariaS [17]3 years ago

4 0

6x -2x^3 + 10x+2x^3+7x^4
combine all like terms to get 7x^4 +16x

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the vertices of triangle PQR are P(-3, 8), Q(-6, -4), and R(1, 1). If you reflect triangle PQR across the x-axis, what will be t

otez555 [7]

Answer:

The coordinates of the vertices of the image will be:

P'=(-3,-8)

Q'=(-6,4)

R'=(1,-1)

Step-by-step explanation:

If you reflect a point P=(x,y) across the x-axis, the coordinates of the image will be P'=(x,-y), then:

P=(-3,8)=(xp,yp)→xp=-3, yp=8→P'=(xp,-yp)→P'=(-3,-8)

Q=(-6,-4)=(xq,yq)→xq=-6, yq=-4→Q'=(xq,-yq)=(-6,-(-4))→Q'=(-6,4)

R=(1,1)=(xr,yr)→xr=1, yr=1→R'=(xr,-yr)→R'=(1,-1)

3 0

4 years ago

Evaluate 3c3 - 7d when c = -4 and d = 6

sergey [27]

Step-by-step explanation:

3c³ - 7d

= 3(-4)³ - 7(6)

= 3(-64) - 42

= -234.

3 0

3 years ago

I’ll give you brainliest for the first person with an answer

raketka [301]

Answer:

1296mm^2

Step-by-step explanation:

Surface Area of the Rectangular Prism to the left on top of the box:

Area of square = lw, 9mm * 9mm = 81mm, then multiply by 2 because 2 of the squares are a part of the surface, giving 162mm^2

Area of a rectangle =lw, 9mm * 9mm = 81mm, then multiply by 2 again, because 2 rectangles are a part of the surface, giving 162mm^2

So, the total surface area of the rectangular prism to the left on top, is 324mm^2

Surface Area of the Triangular Prism:

Area of a triangle: 1/2bh, and our base length would be 12, because we have to subtract 9 from 21 since the base length of the triangle isn't stated. Anyways, A = 1/2bh, so A = 1/2(12)(9) = 54mm^2, but multiply by two, so we get 108mm^2.

Area of the rectangle: lw, so 15mm * 9mm = 135mm^2

So, the total surface area of the triangular prism is 243mm^2

Surface Area of the Rectangular Prism at the bottom:

Area of the long rectangles in front = lw, 21mm * 9mm = 189mm^2, multiply by 2, 378mm^2

Area of the rectangles to the side = lw, 9mm * 9mm = 81mm^2, multiply by 2, 162mm^2

Area of the rectangle at the very bottom = lw, 21mm * 9mm = 189mm^2

So, the total surface area of the rectangular prism at the bottom is 729mm^2

Add all the total surface areas of each shape to get the total surface area of the figure:

324mm^2 + 243mm^2 + 729mm^2 = 1296mm^2

The surface area of the figure above is 1296mm^2

4 0

2 years ago

Shauna makes 24 cupcakes per hour. She has already made 48 cupcakes. Let x represent the amount of time Shauna spends making the

GalinKa [24]

Y=24x+48

24 per hour (x) plus the 48 she already made would equal a total (y).

3 0

3 years ago

The cost of 5 gallons of ice cream has a variance of 49 with a mean of 37 dollars during the summer. What is the probability tha

VMariaS [17]

Answer:

0.0119 = 1.19% probability that the sample mean would be less than 35.2 dollars if a sample of 77 5-gallon pails is randomly selected

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation(which is the square root of the variance) $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .