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3 years ago

What is the value of the number one in 61

Mathematics

2 answers:

ikadub [295]3 years ago

5 0

<span>the value of the number one in 61 is 1
hope it helps</span>

strojnjashka [21]3 years ago

5 0

The value of the number one in 61 is one

hope this helps

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Solve for x: <br> 3x+2y=7

iVinArrow [24]

$3x+2y=7\ \ \ |subract\ 2y\ from\ both\ sides\\\\3x=7-2y\ \ \ |divide\ both\ sides\ by\ 3\\\\\boxed{x=\dfrac{7-2y}{3}}$

8 0

3 years ago

A hockey season ticket holder pays $99.00 for her tickets plus $3.00 for a program for each game. A second person pays $14.00 fo

ddd [48]

The two person will pay the same amount of money when they both watch 9 hockey games.

<h3>How to find the equation and use it to find cost?</h3>

A hockey season ticket holder pays $99.00 for her tickets plus $3.00 for a program for each game. Therefore,

y = 99 + 3x

where

y = total cost

x = number of game

A second person pays $14.00 for a ticket to every game but doesn't buy programs. Therefore,

y = 14x

The number of games they will have the same amount is as follows:

99 + 3x = 14x

99 = 14x - 3x

99 = 11x

x = 99 / 11

x = 9

Therefore, they will have the same amount when they watch 9 games.

learn more on equation here: brainly.com/question/14058936

4 0

2 years ago

9dx10=240 pls hurry

vaieri [72.5K]

Answer:

2.6 or $\frac{8}{3}$

Step-by-step explanation:

9d×10=240

÷10 ÷10

9d =24

÷9 ÷9

2.6 or $\frac{8}{3}$

8 0

2 years ago

judith's family went on a cross-country trip. they drove exactly 232 miles a day for 12 days. how many miles did they drive?

Maksim231197 [3]

Answer:

2784

Step-by-step explanation:

232x12

7 0

2 years ago

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

$\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}$

5 0

3 years ago