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Ksenya-84 [330]

3 years ago

7

Jay had to pain part of the outside of this house he spend 7 hours painting one side and 12 hours painting another how long did

it take him to pain both side?

2 answers:

Whitepunk [10]3 years ago

8 0

Answer:

the answer is 19 hours

stich3 [128]3 years ago

7 0

Answer:

19 hours

Step-by-step explanation:

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Please help!! And btw the little mark by number 6 isn’t a negative, it’s an eraser shaving

____ [38]

For these u can use a caculator for all pretty much.

5 0

2 years ago

Another type of subtraction equation is 16-b=7. Explain how you would sole this equation then solve it.

sladkih [1.3K]

Answer: B = 9

Step-by-step explanation:

16-b=7

Rearrange so we can subtract 16 by 7

16 - 7 = 9

Substituting 9 for b and making sure it's correct.

16 - 9 = 7

4 0

3 years ago

Read 2 more answers

7.15z + 2.3z = ?z i hate math

Nitella [24]

Answer:

9.45z :)

Step-by-step explanation:

3 0

2 years ago

What is an equation of the line that passes through the points (-6, -5) and (-3,0)? Put your answer in fully reduced form.

MrMuchimi

Answer:

y = 5/3x +5

Step-by-step explanation:

The two-point form of the equation of a line is useful here.

y = (y2 -y1)/(x2 -x1)(x -x1) +y1

y = (0 -(-5))/(-3 -(-6))(x -(-6)) -5

y = 5/3(x +6) -5

y = 5/3x +5 . . . . slope-intercept form

5x -3y = -15 . . . .standard form

_____

There is no "fully reduced" form of the equation of a line. There are slope-intercept form, point-slope form, two-point form, standard form, general form, intercept form, and some others. We assume that "fully reduced" applies to any fractions in the equation. The "slope-intercept" form has a fraction, so perhaps that's the form that is required.

6 0

2 years ago

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form ModifyingBelow lim With h

Gre4nikov [31]

Answer:

$\dfrac{1}{2\sqrt{x}}$

Step-by-step explanation:

$f(x) = \sqrt{x} = x^{\frac{1}{2}}$

$f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}$

We use binomial expansion for $(x+h)^{\frac{1}{2}}$

This can be rewritten as

$[x(1+\dfrac{h}{x})]^{\frac{1}{2}}$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}$

From the expansion

$(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots$

Setting $x=\dfrac{h}{x}$ and $n=\frac{1}{2}$ ,

$(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots$

$=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots$

Multiplying by $x^{\frac{1}{2}}$ ,

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots$

The limit of this as $h\to 0$ is

$\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}}$ (since all the other terms involve $h$ and vanish to 0.)

8 0

3 years ago