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3 years ago

What is 5h-9= -16+6h

2 answers:

6 0

$5h - 9 -16 + 6h \\ \\ 5h - 9 = 6h - 16 \\ \\ -9 = 6h - 16 - 5h \\ \\ -9 = h - 16 \\ \\ -9 + 16 = h \\ \\ 7 = h \\ \\ h = 7 \\ \\ Answer: \fbox {h = 7}$

valkas [14]3 years ago

4 0

5h - 9 = -16 + 6h

Combine like terms

5h subtracted from both sides.

-9+16 = -16 + h
+16

-9+16 = 7

h = 7

Answer: h = 7

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What is the distance from L to M?

ANEK [815]

Answer:

A. 10 units

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)
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Algebra II

Distance Formula: $\displaystyle d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Step-by-step explanation:

Step 1: Define

Point L(-5, 4)

Point M(3, -2)

Step 2: Find distance d

Simply plug in the 2 coordinates into the distance formula to find distance d

Substitute in points [Distance Formula]: $\displaystyle d = \sqrt{(3+5)^2+(-2-4)^2}$
[√Radical] (Parenthesis} Add/Subtract: $\displaystyle d = \sqrt{(8)^2+(-6)^2}$
[√Radical] Evaluate exponents: $\displaystyle d = \sqrt{64+36}$
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8 0

3 years ago

Again.. your girl needs help again ..

Maru [420]

Answer: C. $3x^{2}$ + $3$

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4 years ago

Mt. Everest is 8,848 meters tall. If there

sergeinik [125]

Answer:29,198.4

Step-by-step explanation:

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5 0

3 years ago

The star Sirius is more than 4 light-years farther than earth than the star Centauri. Write an inequality that represents the di

vredina [299]

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3 years ago

Find the differential of each function. (a) y = tan( 5t ) dy = Correct: Your answer is correct. (b) y = 5 − v2 5 + v2

Trava [24]

Answer:

$(a)$ $dy = 5sec^2(5t) \ dt$

$(a) \ dy = \frac{-20v}{(5+v^2)^2} \ dt$

Step-by-step explanation:

Given;

(a) y = tan(5t)

$(b) \ y = \frac{5-v^2}{5+v^2}$

Solving for (a)

y = tan(5t)

let u = 5t

⇒y = tan(u)

du/dt = 5

dy/du = sec²u

$\frac{dy}{dt} =\frac{dy}{du} *\frac{du}{dt} \\\\\frac{dy}{dt} = sec^2(u)*5\\\\\frac{dy}{dt} = 5sec^2(u)\\\\\frac{dy}{dt} = 5sec^2(5t)$

$dy = 5sec^2(5t) \ dt$

Solving for b;

let u = 5 - v²

du/dv = -2v

let v = 5+ v²

dv/du = 2v

$\frac{dy}{dv} = \frac{vdu - udv}{v^2} \\\\\frac{dy}{dv} = \frac{-2v(5+v^2) - 2v(5-v^2)}{(5+v^2)^2}\\\\\frac{dy}{dv} = \frac{-10v-2v^3-10v+2v^3}{(5+v^2)^2}\\\\$

$\frac{dy}{dv} = \frac{-10v-10v}{(5+v^2)^2}\\\\\frac{dy}{dv} = \frac{-20v}{(5+v^2)^2}\\\\dy = \frac{-20v}{(5+v^2)^2} dt$

4 0

3 years ago