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Brums [2.3K]
3 years ago
7

G=qd+hd, solve for d

Mathematics
2 answers:
Bad White [126]3 years ago
7 0
Remember distributive property
ab+ac=a(b+c)
undistribute d
G=d(q+h)
divide both sides by (q+h)
\frac{G}{q+h} =d
Lyrx [107]3 years ago
5 0
Let's solve for d:
g=qd+hd
Step 1: Flip the equation.
dh+dq=g
Step 2: Factor out variable d.
d(h+q)=g
Step 3: Divide both sides by by h+q
d(h+q)/h+q=g/h+q
Answer:
d=g/h+q
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Olegator [25]
Answer is C because 0.3 is not positive number or even negative number.
4 0
3 years ago
Read the power and then check all that apply.
erastovalidia [21]
Check all the ones listed below:

1. The base is 3

(you are multiplying 3 four times or 3 to the power of 4 )

2. The exponent is 4

( 3 is the number you are multiplying and the exponent or 4 tells you how many times you multiply the base which is 3)

3. The exponent tells you to multiply the 3 together 4 times

( 3x3x3x3 is multiplying the number 3, 4 times)
6 0
3 years ago
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PLZ HELP ASAP!!!!!!!!!!!!!!!!!
eimsori [14]
B and C because both bases are squares
8 0
3 years ago
Math help Please!!!!
GalinKa [24]

Part A. What is the slope of a line that is perpendicular to a line whose equation is −2y=3x+7?

Rewrite the equation  −2y=3x+7 in the form y=-\dfrac{3}{2}x-\dfrac{7}{2}. Here the slope of the given line is  m_1=-\dfrac{3}{2}. If m_2 is the slope of perpendicular line, then

m_1\cdot m_2=-1,\\ \\m_2=-\dfrac{1}{m_1}=\dfrac{2}{3}.

Answer 1: \dfrac{2}{3}

Part B. The slope of the line y=−2x+3 is -2. Since -\dfrac{3}{2}\neq -2\quad \text{and}\quad \dfrac{2}{3}\neq -2, then lines from part A are not parallel to line a.

Since -2\cdot \left(-\dfrac{3}{2}\right)=3\neq -1\quad \text{and}\quad -2\cdot \dfrac{2}{3}=-\dfrac{4}{3}\neq -1, both lines are not perpendicular to line a.

Answer 2: Neither parallel nor perpendicular to line a

Part C. The line parallel to the line 2x+5y=10 has the equation 2x+5y=b. This line passes through the point (5,-4), then

2·5+5·(-4)=b,

10-20=b,

b=-10.

Answer 3: 2x+5y=-10.

Part D. The slope of the line y=\dfrac{x}{4}+5 is \dfrac{1}{4}. Then the slope of perpendicular line is -4 and the equation of the perpendicular line is y=-4x+b. This line passes through the point (2,7), then

7=-4·2+b,

b=7+8,

b=15.

Answer 4: y=-4x+15.

Part E. Consider vectors \vec{p}_1=(-c-0,0-(-d))=(-c,d)\quad \text{and}\quad \vec{p}_2=(0-b,a-0)=(-b,a). These vectors are collinear, then

\dfrac{-c}{-b}=\dfrac{d}{a},\quad \text{or}\quad -\dfrac{a}{b}=-\dfrac{d}{c}.

Answer 5: -\dfrac{a}{b}=-\dfrac{d}{c}.

5 0
4 years ago
Read 2 more answers
Trigonometry Identities help please
sesenic [268]

theta is in the fourth quadrant where the cosine is positive.


the third side in the triangle = sqrt (4 - 2) = sqrt2


So sin theta = -sqrt2/2 = Second choice (negative because sine is negative in 4th quadrant)


tan theta = - sqrt2 / sqrt2 = -1

6 0
3 years ago
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