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gizmo_the_mogwai [7]

3 years ago

13

A translation moves point V(-2,3) to V’(2,7). Which are true statements about the translation?

1 answer:

Stolb23 [73]3 years ago

3 0

Answer:

The translation is right 4 units and up 4 units

Step-by-step explanation:

we know that

The rule of the translation of the point V to V' is equal to

(x,y) ----> (x+4,y+4)

That means ----> The translation is right 4 units and up 4 units

Verify

V(-2,3) -----> V'(-2+4,3+4)

V(-2,3) -----> V'(2,7)

Is correct

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Dan uses a remote control to make his drone take off from the ground. Function h represents the height, in feet, of the drone t

LUCKY_DIMON [66]

I think it’s c I am not sure

3 0

3 years ago

Read 2 more answers

Simplify<br> 7.1x106<br> ______<br> 8.2 x 101

vekshin1

Answer:

7.1x106=752.6

8.2 x 101=828.2

3 0

2 years ago

For what value of k is there one solution to the given quadratic equation (k+1)x²+4kx+2=0.

andriy [413]

Answer:

If $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

The signal of $\bigtriangleup$ determines how many real roots an equation has:

$\bigtriangleup > 0$ : Two real and different solutions

$\bigtriangleup = 0$ : One real solution

$\bigtriangleup < 0$ : No real solutions

In this problem, we have the following second order polynomial:

$(k+1)x^{2} + 4kx + 2 = 0$ .

This means that $a = k+1; b = 4k; c = 2$

It has one solution if

$\bigtriangleup = 0$

$b^{2} - 4ac = 0$

$16k^{2} -8(k+1) = 0$

$16k^{2} - 8k - 8 = 0$

We can simplify by 8

$2k^{2} - k - 1 = 0$

The solution is:

$k = 1$ or $k = -\frac{1}{2}$

So, if $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

4 0

3 years ago

In the circle below, DB = 22 cm, and m<DBC = 60°. Find BC. Ignore my handwriting.

Karo-lina-s [1.5K]

Answer:

$BC=11\ cm$

Step-by-step explanation:

step 1

Find the measure of the arc DC

we know that

The inscribed angle measures half of the arc comprising

$m\angle DBC=\frac{1}{2}[arc\ DC]$

substitute the values

$60\°=\frac{1}{2}[arc\ DC]$

$120\°=arc\ DC$

$arc\ DC=120\°$

step 2

Find the measure of arc BC

we know that

$arc\ DC+arc\ BC=180\°$ ----> because the diameter BD divide the circle into two equal parts

$120\°+arc\ BC=180\°$

$arc\ BC=180\°-120\°=60\°$

step 3

Find the measure of angle BDC

we know that

The inscribed angle measures half of the arc comprising

$m\angle BDC=\frac{1}{2}[arc\ BC]$

substitute the values

$m\angle BDC=\frac{1}{2}[60\°]$

$m\angle BDC=30\°$

therefore

The triangle DBC is a right triangle ---> 60°-30°-90°

step 4

Find the measure of BC

we know that

In the right triangle DBC

$sin(\angle BDC)=BC/BD$

$BC=(BD)sin(\angle BDC)$

substitute the values

$BC=(22)sin(30\°)=11\ cm$

6 0

3 years ago

Solve for x:<br> 1 1/5x−2 1/3<1/7x+1 1/2

Mashcka [7]

Answer: $x$

Step-by-step explanation:

$1\frac{1}{5}x-2\frac{1}{3}$

Add $2\frac{1}{3}$ on both sides and subtract $\frac{1}{7}x$ on both sides to leave x's on the left side and independent values on the right.

$(2\frac{1}{3}-\frac{1}{7}x) +1\frac{1}{5}x-2\frac{1}{3}$

$1\frac{1}{5}x-\frac{1}{7}x$

Solve the fractions.

$1(\frac{1}{5}-\frac{1}{7})x$

$1(\frac{(1)(7)-(5)(1)}{(5)(7)} )x$

$1(\frac{7-5}{35} )x$

$1(\frac{2}{35} )x$

$1\frac{2}{35}x$

Convert the mixed fraction $1\frac{2}{35}$ to an improper fraction. You can do this by multiplying 1 times 35 and adding 2.

$\frac{37}{35}x$

Now use the reciprocal (inverse fraction) and multiply on both sides to isolate x.

$(\frac{35}{37}) (\frac{37}{35})x$

$x$

$x$

$x$

5 0

3 years ago