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3 years ago

????????????????????????

Mathematics

1 answer:

lakkis [162]3 years ago

5 0

The parallelogram ABCD is translated 7 units to the left to create A'B'C'D'.

Hope this helps :)

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How do you solve this in standard form. y=7x-4

attashe74 [19]

Y = 2/7x - 4
y + 4 = 2/7x
4 = 2/7x - y (multiply everything by seven to get rid of the fraction)
28 = 2x - 7y
2x - 7y = 28

8 0

3 years ago

A pair of reading glasses shows you an image of an object that is 3.1 times the object's actual size. So the scale factor of the

KiRa [710]

When given the scale factor and the factored size of the object, to find the original size, divide the factored size by the scale factor:

0.8 / 3.1 = 0.258 inches

3 0

3 years ago

PLEASE ANSWER ASAP WORTH 100 points get it right please

Trava [24]

Answer:

See below.

Step-by-step explanation:

Part A:

triangle QPR and triangle PSR are similar

Part B:

Triangle QPR is a right triangle with right angle QPR.

Triangle SPR is a right triangle with right angle PSR.

Angle QPR of triangle QPR corresponds to and is congruent to angle PSR of triangle PSR.

Angle R of triangle QSR corresponds to and is congruent to angle R of triangle PSR.

The similar triangles by AA Similarity are

triangle QPR and triangle PSR

Part C:

QR/PR = PR/SR

16/PR = PR/4

PR² = 16 * 4

PR² = 64

PR = 8

8 0

2 years ago

Is the quotient of two irrational numbers sometime, always, or never irrational number?

Varvara68 [4.7K]

Sometimes, because if you divide irrational numbers by itself it'll just equal 1, otherwise you would get an irrational number.

7 0

3 years ago

What numbers are zeros of g(x) = x^2 - 2x - 4? take your time if you need to!

ipn [44]

Answer:

$x = 1 + \sqrt5, x = 1 - \sqrt5$

Step-by-step explanation:

Hello!

We can solve the quadratic by using the quadratic formula.

Standard form of a quadratic: $ax^2 + bx + c = 0$

Quadratic Formula: $x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$

Given our Equation: $g(x) = x^2 - 2x - 4$

a = 1
b = -2
c = -4

Plug the values into the equation and solve.

<h3>Solve</h3>

$x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$
$x = \frac{-(-2)\pm\sqrt{(-2)^2 - 4(1)(-4)}}{2(1)}$
$x = \frac{2\pm\sqrt{4 +16}}{2}$
$x = \frac{2\pm\sqrt{20}}{2}$
$x = \frac{2\pm\sqrt{4 * 5}}{2}$
$x = \frac{2\pm(\sqrt4 * \sqrt5)}{2}$
$x = \frac{2\pm2\sqrt5}{2}$
$x = 1 + \sqrt5, x = 1 - \sqrt5$

7 0

2 years ago