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3 years ago

9

H+3h+4h=100 how do i get h?

2 answers:

Dominik [7]3 years ago

7 0

Answer:

h = 12.5

Step-by-step explanation:

Simplifying

h + 3h + 4h = 100

Combine like terms: h + 3h = 4h

4h + 4h = 100

Combine like terms: 4h + 4h = 8h

8h = 100

Solving

8h = 100

Solving for variable 'h'.

Move all terms containing h to the left, all other terms to the right.

Divide each side by '8'.

h = 12.5

Simplifying

h = 12.5

rewona [7]3 years ago

6 0

Answer:

h=12.5

Step-by-step explanation:

To solve for h, we need to isolate the variable

h+3h+4h =100

Combine like terms by adding h, 3h and 4h

8h=100

Divide both sides by 8 to "reverse" the multiplication being done to h

8h/8=100/8

The 8s on the left will cancel, leaving us with:

h=12.5

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3 years ago

1/5+1/7+3/21 the answer is

inessss [21]

Answer:

<h2>17/35</h2>

Step-by-step explanation:

We need to get all the fractions with a common denominator before we can solve the problem.

The only fractions that have a common multiple are 1/7 and 3/21, so we'll add these first and then add 1/5.

1/7 = 3/21

Add

$\frac{3}{21} + \frac{3}{21} =\frac{6}{21}$

6/21 = 30/105

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I'm always happy to help :)

4 0

2 years ago

g Use this to find the equation of the tangent line to the parabola y = 2 x 2 − 7 x + 6 at the point ( 4 , 10 ) . The equation o

natali 33 [55]

Answer:

The tangent line to the given curve at the given point is $y=9x-26$ .

Step-by-step explanation:

To find the slope of the tangent line we to compute the derivative of $y=2x^2-7x+6$ and then evaluate it for $x=4$ .

$(y=2x^2-7x+6)'$ Differentiate the equation.

$(y)'=(2x^2-7x+6)'$ Differentiate both sides.

$y'=(2x^2)'-(7x)'+(6)'$ Sum/Difference rule applied: $(f(x)\pmg(x))'=f'(x)\pm g'(x)$

$y'=2(x^2)'-7(x)'+(6)'$ Constant multiple rule applied: $(cf)'=c(f)'$

$y'2(2x)-7(1)+(6)'$ Applied power rule: $(x^n)'=nx^{n-1}$

$y'=4x-7+0$ Simplifying and apply constant rule: $(c)'=0$

$y'=4x-7$ Simplify.

Evaluate y' for x=4:

$y'=4(4)-7$

$y'=16-7$

$y'=9$ is the slope of the tangent line.

Point slope form of a line is:

$y-y_1=m(x-x_1)$

where $m$ is the slope and $(x_1,y_1)$ is a point on the line.

Insert 9 for $m$ and (4,10) for $(x_1,y_1)$ :

$y-10=9(x-4)$

The intended form is $y=mx+b$ which means we are going need to distribute and solve for $y$ .

Distribute:

$y-10=9x-36$

Add 10 on both sides:

$y=9x-26$

The tangent line to the given curve at the given point is $y=9x-26$ .

------------Formal Definition of Derivative----------------

The following limit will give us the derivative of the function $f(x)=2x^2-7x+6$ at $x=4$ (the slope of the tangent line at $x=4$ ):

$\lim_{x \rightarrow 4}\frac{f(x)-f(4)}{x-4}$

$\lim_{x \rightarrow 4}\frac{2x^2-7x+6-10}{x-4}$ We are given f(4)=10.

$\lim_{x \rightarrow 4}\frac{2x^2-7x-4}{x-4}$

Let's see if we can factor the top so we can cancel a pair of common factors from top and bottom to get rid of the x-4 on bottom:

$2x^2-7x-4=(x-4)(2x+1)$

Let's check this with FOIL:

First: $x(2x)=2x^2$

Outer: $x(1)=x$

Inner: $(-4)(2x)=-8x$

Last: $-4(1)=-4$

---------------------------------Add!

$2x^2-7x-4$

So the numerator and the denominator do contain a common factor.

This means we have this so far in the simplifying of the above limit:

$\lim_{x \rightarrow 4}\frac{2x^2-7x-4}{x-4}$

$\lim_{x \rightarrow 4}\frac{(x-4)(2x+1)}{x-4}$

$\lim_{x \rightarrow 4}(2x+1)$

Now we get to replace x with 4 since we have no division by 0 to worry about:

2(4)+1=8+1=9.

6 0

3 years ago