Which rule describes the composition oftransformations that maps ABC to A'B'C"?

Mathematics

2 answers:

krek1111 [17]3 years ago

8 0

Answer:

translation

Step-by-step explanation:

Sophie [7]3 years ago

3 0

I think it’s transfiguration :)

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Which statement about rhombuses and parallelograms is true?

puteri [66]

Answer:

Parallelograms ALWAYS have opposite sides parallel; rhombuses sometimes do not.

Step-by-step explanation:

i think

4 0

4 years ago

7x^2=9+x what are the values of x Will give medal and points

DENIUS [597]

7x² = 9 + x Subtract x from both sides
7x² - x = 9 Subtract 9 from both sides
7x² - x - 9 = 0 Use the Quadratic Formula

a = 7 , b = -1 , c = -9

x = $\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$ Plug in the a, b, and c values
x = $\frac{- (-1) \pm \sqrt{(-1)^2 - 4(7)(-9)} }{2(7)}$ Cancel out the double negative
x = $\frac{1 \pm \sqrt{(-1)^2 - 4(7)(-9)} }{2(7)}$ Square -1
x = $\frac{1 \pm \sqrt{1 - 4(7)(-9)} }{2(7)}$ Multiply 7 and -9
x = $\frac{1 \pm \sqrt{1 - 4(-63} }{2(7)}$ Multiply -4 and -63
x = $\frac{1 \pm \sqrt{1 + 252} }{2(7)}$ Multiply 2 and 7
x = $\frac{1 \pm \sqrt{1 + 252} }{14}$ Add 1 and 252
x = $\frac{1 \pm \sqrt{253} }{14}$ Split up the $\pm$
x = $\left \{ {{ \frac{1 + \sqrt{253} }{14} } \atop { \frac{1 - \sqrt{253} }{14} }} \right.$
The approximate square root of 253 is 15.905973.
x ≈ $\left \{ { \frac{1 + 15.905973}{14} } \atop { \frac{1 - 15.905973}{14} }} \right$ Add and subtract
x ≈ $\left \{ {{ \frac{16.905973}{14} } \atop { \frac{14.905973}{14} }} \right.$ Divide
x ≈ $\left \{ {{1.2075} \atop {1.0647}} \right.$ Round to the nearest hundredth
x ≈ $\left \{ {{1.21} \atop {1.06}} \right.$

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3 years ago

Simplify the expression 5+4+2(-3)+7+(-5)

liberstina [14]

Just use pemdas
(Parenthesis, exponents, multiplication, division, addition, subtraction)

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3 years ago

Find the equation of a line that passes the points (5, -9) and (0, -3)

kobusy [5.1K]

Answer:

y = -6/5x - 3

Step-by-step explanation: