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3 years ago

Find the midpoint of the line segment given the endpoints. (-1, 10) (-1, 6)

Mathematics

1 answer:

BaLLatris [955]3 years ago

8 0

The midpoint of these two points is (-1, 8).

In order to find the midpoint, you must take the average of each part of the coordinates. You do this by adding them together and dividing by 2. We'll start with the x-coordinates.

(-1 + -1)/2 = -1

Now we'll use the y-coordinates.

(10 + 6)/2 = 8

This gives us the ordered pair of (-1, 8)

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What is the height of this isosceles triangle?

erastova [34]

Answer:

height ≈ 11.9 units

Step-by-step explanation:

Construct a perpendicular bisector from the vertex to the base, thus splitting the triangle into 2 right triangles.

Using the sine ratio in one of these right triangles, then

sin48° = $\frac{opposite}{hypotenuse}$ ( the opposite is the height h )

sin48° = $\frac{h}{16}$ ( multiply both sides by 16 )

16 × sin48° = h , thus

h ≈ 11.9 ( to the nearest tenth )

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3 years ago

H(x) = x -2<br> g(x) = 2x² – 3x<br> Find h(g(-2))

emmasim [6.3K]

Answer:

h[g(-2)] = 12

Step-by-step explanation:

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3 years ago

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Find the median of the set of data. 30, 16, 49, 25 <br><br>

weeeeeb [17]

Answer:

16,25,30,49

25+30=55

55÷2=27.5

Step-by-step explanation:

median means the middle value of any set of data so first arrange the data into ascending order.

16, 25, 30, 49

the data is even so we take both the middle value , add it and divide it with 2

25+30=55

55÷2=27.5

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3 years ago

43 ones x 3 ten equals

Svet_ta [14]

The answer is 1230!!!!!!!

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3 years ago

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What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .

Subtract $-12x^2 -36x$ from both sides of the equation:

$38x + 7 = c(x + 3)$ .

By the same logic, $c = 38$ .

Hence the quotient would be $(4x^2 - 12x + 38)$ .

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3 years ago