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3 years ago

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help only one question please with full step amd if you are right i will mark you as brainlist. 30 point

1 answer:

Tom [10]3 years ago

6 0

Answer:

Step-by-step explanation:

The scale factor describes the size of an enlargement or reduction. For example, a scale factor of means that the new shape is twice the size of the original. A scale factor of means that the new shape is three times the size of the original.

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Y^2-2y+1=x+5<br><br> Make y the theme of the formula in the equation

Tanzania [10]

Answer: y = 1 +/- $\sqrt{x+5}$

<u>Step-by-step explanation:</u>

y² - 2y + 1 = x + 5

(y - 1)² = x + 5

$\sqrt{(y-1)^{2}} = \sqrt{x+5}$

y - 1 = +/- $\sqrt{x+5}$

y = 1 +/- $\sqrt{x+5}$

7 0

3 years ago

How do you find the slope of a line perpendicular to a given line?

enyata [817]

You have a line:
y=mx+b (slope-intercepted form)
m=slope of this line.

The slope of a line perpendicular to that given line will be: ´"m´"
m´=-1/m.

For example:
y=8x+3
m=8

The solpe fo a line perpendicular to "y=8x+3" is:
m`=-1/8

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%20%5Csf%20%5Chuge%7B%20question%20%5Chookleftarrow%7D" id="TexFormula1" title=" \sf \huge

BabaBlast [244]

$\underline{\bf{Given \:equation:-}}$

$\\ \sf{:}\dashrightarrow ax^2+by+c=0$

$\sf Let\:roots\;of\:the\: equation\:be\:\alpha\:and\beta.$

$\sf We\:know,$

$\boxed{\sf sum\:of\:roots=\alpha+\beta=\dfrac{-b}{a}}$

$\boxed{\sf Product\:of\:roots=\alpha\beta=\dfrac{c}{a}}$

$\underline{\large{\bf Identities\:used:-}}$

$\boxed{\sf (a+b)^2=a^2+2ab+b^2}$

$\boxed{\sf (√a)^2=a}$

$\boxed{\sf \sqrt{a}\sqrt{b}=\sqrt{ab}}$

$\boxed{\sf \sqrt{\sqrt{a}}=a}$

$\underline{\bf Final\: Solution:-}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}$

$\bull\sf Apply\: Squares$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2= (\sqrt{\alpha})^2+2\sqrt{\alpha}\sqrt{\beta}+(\sqrt{\beta})^2$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2 \alpha+\beta+2\sqrt{\alpha\beta}$

$\bull\sf Put\:values$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2=\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\sqrt{\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}}$

$\bull\sf Simplify$

$\\ \sf{:}\dashrightarrow \underline{\boxed{\bf {\sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\sqrt{\dfrac{-b}{a}}+\sqrt{2}\dfrac{c}{a}}}}$

$\underline{\bf More\: simplification:-}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{-b}}{\sqrt{a}}+\dfrac{c\sqrt{2}}{a}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{a}\sqrt{-b}+c\sqrt{2}}{a}$

$\underline{\Large{\bf Simplified\: Answer:-}}$

$\\ \sf{:}\dashrightarrow\underline{\boxed{\bf{ \sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\dfrac{\sqrt{-ab}+c\sqrt{2}}{a}}}}$

5 0

2 years ago

Read 2 more answers

Is this question a statistical question or a non statistical question : If I

enot [183]

Answer:non statistical

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The ordered pair (-4, -5) is a solution of the following system of equations 2x+y=13 19x+13y=15

liberstina [14]

Answer:

Step-by-step explanation

672

8 0

3 years ago