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2 years ago

A contractor is building a fence around a square garden that has a side length of 3 inches in a

Mathematics

1 answer:

Pavel [41]2 years ago

7 0

Answer:

is it adding or subtraction

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For all functions of the form f(x) = ax2 + bx + c, which is true when b = 0?

Klio2033 [76]

When b=0
f(x)=ax^2+c

test each
A. with x^2-1, A is false
B. with -x^2-1, B is false
C. cannot find contradiction
D. the axis is actually x=0, 0 is nithere positive nor negative, false

answer is C

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4 years ago

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Please i need help !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

tekilochka [14]

Answer:

limeted government or rule of law i think sorta mayby kindaa

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3 years ago

How many kilograms are 0.54 decalitres?

Anarel [89]

Answer:

A measure of capacity or volume in the metric system equal to one-tenth (10-1) of a liter. One deciliter is also equal to 6.1022 cubic inches, or 3.38 fluid ounces

Step-by-step explanation:

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3 years ago

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XYZ has coordinates X(2, 3), Y(1,4), and Z(8,9). A translation maps X to X'(4,7). What are the coordinates for Y' and Z' for thi

Rashid [163]

Answer:

The coordinates are $Y'(x,y) = (3, 8)$ and $Z'(x,y) = (10, 13)$ .

Step-by-step explanation:

First, we have to derive an expression for translation under the assumption that each point of XYZ experiments the same translation. Vectorially speaking, translation from X to X' is defined by:

$X'(x,y) = X(x,y) + T(x,y)$ (1)

Where $T(x,y)$ is the vector translation.

If we know that $X(x,y) = (2,3)$ and $X'(x,y) = (4,7)$ , then the vector translation is:

$T(x,y) = X'(x,y)-X(x,y)$

$T(x,y) = (4,7) - (2,3)$

$T(x,y) = (2, 4)$

Then, we determine the coordinates for Y' and Z':

$Y'(x,y) = Y(x,y) + T(x,y)$

$Y'(x,y) = (1,4) + (2,4)$

$Y'(x,y) = (3, 8)$

$Z'(x,y) = Z(x,y) + T(x,y)$

$Z'(x,y) =(8,9) + (2,4)$

$Z'(x,y) = (10, 13)$

The coordinates are $Y'(x,y) = (3, 8)$ and $Z'(x,y) = (10, 13)$ .

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3 years ago

Please solve all with explanation

garri49 [273]

Answer:

1. $\frac{x^2}{y^8}$

2 a) $\frac{-49}{2}$

b) 0

3. x = 10

Step-by-step explanation:

$(\frac{x^{-3}y^4}{x^{-2}})^{-2}\\({x^{-3+2}y^4})^{-2}\\({x^{-1}y^4})^{-2}({x^{-1*-2}y^{4*-2})= \frac{x^2}{y^8}$

2. a)

$2 [(\frac{-1}{4}) -12 (-1)^2] - 5[24 +2 (-12)]^2\\2[(\frac{-1}{4}) -12] - 5[24 -24]^2\\2 * [(\frac{-1}{4}) -12] -0\\\frac{-49}{2}$

$-a^2b^2 - 3a^3b^2\\-\frac{1}{9}*1 - 3 * \frac{-1}{27} *1\\-\frac{1}{9} + \frac{1}{9}\\0$

$\frac{3}{4}x + \frac{5}{6} = 5x -\frac{125}{3} \\\frac{5}{6} + \frac{125}{3} = 5x - \frac{3}{4}x\\\frac{5+250}{6} = \frac{20x -3x}{4}\\\frac{255}{6} = \frac{17x}{4}\\x = 10$

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3 years ago