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3 years ago

There are 300 pages in your independent reading book for English. You have read 78% of the book. How many pages have you read?

Mathematics

2 answers:

mario62 [17]3 years ago

8 0

Answer:

234

Step-by-step explanation:

300 x .78 = 234

Juliette [100K]3 years ago

7 0

Answer:

234 pages.

Step-by-step explanation:

Remark

78% = 78/100

78% of 300 = 78/100 * 300 = 23400/100 = 234

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Answer:

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Step-by-step explanation:

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3 years ago

Solve the following system of equations using Substitution: Submit your answer as an ordered pair. 2x−3y=32x-3y=32x−3y=3 x=y+2x=

exis [7]

9514 1404 393

Answer:

(3, 1)

Step-by-step explanation:

We assume you want the solution to the system ...

2x−3y=3
x=y+2

The second equation gives a nice expression for x, so we can use that in the first equation.

2(y+2) -3y = 3 . . . . substitute for x in the first equation

2y +4 -3y = 3 . . . . . eliminate parentheses

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4 0

2 years ago

4 + 6d = 34<br> Need help solving

stealth61 [152]

Answer:

order of operation

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then it is 6d=30

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ANSWER=5

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Step-by-step explanation:

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2 years ago

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The point(5/13,Y) IS IN THE 4TH quadrant corresponds to angle on the unit circle

shutvik [7]

4th quadrant is positive x (cos) and negative y (sin)

Use the Pythagorean Theorem to calculate the value of y (sin).
x² + y² = c²
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25 + y² = 169
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Since the y-value is negative in the 4th quadrant, then y = -12

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3 years ago

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BabaBlast [244]

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

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

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

$\sf We\:know,$

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

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

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

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

$\boxed{\sf (√a)^2=a}$

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

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

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

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

$\bull\sf Apply\: Squares$

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

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

$\bull\sf Put\:values$

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

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

$\bull\sf Simplify$

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

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

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

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

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

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

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2 years ago

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