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2 years ago

Sum of 4x^3+2x^2-4x+3 and 7x^3-4x+7x+8

Mathematics

1 answer:

Flura [38]2 years ago

3 0

Answer:

11x^3−2x^2+3x+11 this is the correct answer

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What is the range of the function y = x² + 1?

Mazyrski [523]

Answer

[ 1 by 2, -1 by 2 ]

Step-by-step explanation:

https://haygot.s3.amazonaws.com/questions/1118559_1200733_ans_2ec06a240d2240e49dc1c9fda7b41ef5.jpeg

7 0

2 years ago

Read 2 more answers

PLEASE HELP!!! <br> WHICH OF THE FOLLOWING FUNCTIONS IS GRAPHED BELOW?

SashulF [63]

Answer: D.

Step-by-step explanation:

For an absolute value function, the vertex of $f(x) = |x+h| + k$ is defined as the point (-h, k) for the coordinate (x, y).

When x is equal to negative h, the value for x and value for h effectively cancel out, and only the positive k remains, hence the vertex being (-h, k).

The function given has a vertex at (2, 3). We know that the vertex of an absolute function is (-h, k), so h must equal -2 and k must equal 3.

The equation:

$f(x) = |x-2| + 3$

7 0

2 years ago

I NEED HELP PLEASE THANKS

Simora [160]

Answer:

D. Reflect across the x-axis, then shift 4 units up

Step-by-step explanation:

In this case, reflecting across the x-axis just means going 4 units down. To go back up, go 4 units up.

8 0

2 years ago

A big lockers above a smaller locker they both have 0.5 m wide and 0.6 m deep Big lockers 1.2 m tall and small lockers half of t

Tomtit [17]

Answer:

$0.54\,\,m^3$

Step-by-step explanation:

Given: Dimensions of a big locker are 0.5 m × 0.6 m × 1.2 m

Dimensions of a small locker are 0.5 m × 0.6 m × $\frac{1.2}{2}=0.6\,\,m$ (as height of small locker is half the height of big locker )

To find: total volume of one big locker and one small locker

Solution:

Volume of cuboid = length × breadth × height

Total volume of one big locker and one small locker = Total volume of one big locker + total volume of one small locker

= $0.5\times 0.6\times 1.2+0.5\times 0.6\times 0.6$

$=0.5\times 0.6\left ( 1.2+0.6 \right )\\=0.3(1.8)\\=0.54\,\,m^3$

4 0

2 years ago

Read 2 more answers

Make r the subject of the formula <br>v = \pi \: h {}^{2}(r - \frac{h}{3})v=πh2(r−3h) <br>

mel-nik [20]

Answer:

$\boxed{r = \frac{h}{3} + \frac{v}{\pi {h}^{2} } }$

Step-by-step explanation:

$Solve \: for \: r: \\ = > v= \pi {h}^{2}(r - \frac{h}{3} ) \\ \\ v=\pi {h}^{2}(r - \frac{h}{3} )is \: equivalent \: to \: {h}^{2}\pi(r - \frac{h}{3} ) = v: \\ = > {h}^{2}\pi(r - \frac{h}{3} ) = v \\ \\ Divide \: both \: sides \: by \: \pi {h}^{2} : \\ = > r - \frac{h}{3} = \frac{v}{\pi {h}^{2} } \\ \\ Add \: \frac{h}{3} \: to \: both \: sides: \\ = > r = \frac{h}{3} + \frac{v}{\pi {h}^{2} }$

7 0

2 years ago