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alukav5142 [94]

3 years ago

5

Dale is riding on a bike course that is miles long. So far, he has ridden miles of the course. What percentage of the course has

Dale ridden so far?

1 answer:

adelina 88 [10]3 years ago

8 0

Answer:

40%

Step-by-step explanation:

Total distance to be covered with the bike ride = 45 miles

Distance covered till now = 18 miles.

Percentage covered = $\frac{distance \\ covered}{Total \\ distance}$

Therefore as a percentage it shall be = $\frac{18}{45} = 0.4$

Since percentages are presented in whole numbers, the result of division shall be multiplied with 100.

Accordingly, total distance covered in percentage = 0.4 $\times$ 100 = 40%

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PLEASE HELPPP<br><br> Which expression is equivalent to

Anettt [7]

Answer:

Fourth option

Step-by-step explanation:

Given the expression in the picture:

Apply Power of Power Rule

$\frac{64g {}^{9} h {}^{6}k {}^{12} }{8g {}^{3} h {}^{2} }$

$8g {}^{6} h {}^{4} k {}^{12}$

The left side of the equation option had this in the fourth option so the fourth option is right

7 0

3 years ago

Please solve and explain the question below

solong [7]

Answer:

OPTION A

Step-by-step explanation:

To find the table substitute the points on the given function and compare the values.

The given function is: $y = x^2 + 2$ .

OPTION A:

(i) When x = -2

LHS = y = 6.

RHS = (-2)² + 2 = 4 + 2 = 6.

LHS = RHS

(ii) When x = -1

LHS = y = 3

RHS = (-1)² + 2 = 1 + 2 = 3.

LHS = RHS

(iii) When x = 0

LHS = y = 2

RHS = 0² + 2 = 2.

LHS = RHS

(iv) When x = 1

LHS = y = 3

RHS = (1)² + 2 = 1 + 2 = 3.

LHS = RHS

(v) When x = 2

LHS = y = 6

RHS = (2)² + 2 = 4 + 2 = 6

LHS = RHS

OPTION B:

(i) When x = -2

LHS = y = -2

RHS = (-2)² + 2 = 6

LHS $\neq$ RHS

OPTION B is eliminated.

OPTION C:

(i) When x = -2

Using the same reason as OPTION B this option is eliminated as well.

So, OPTION A is the correct answer.

5 0

3 years ago

How do you solve this?

puteri [66]

Answer:

Width: 7

Length: 14

Step-by-step explanation:

The area of a rectangle can be found my multiplying the length by the width. We do not know neither the length nor the width. However, we do know that the length is 7 feet longer than the width. Because we do not know the value of either side, let's let $x$ represent the width.

Since the length is 7 feet longer, we can represent the length by $x+7$ .

Now that we have our values for the length and the width we can multiply them together to find our area.

$(x)(x+7)=x^{2} +7x$

Now we have our equation, so we can address the changes made by the original question. The original question states that when 7 is added to both the length and width the area becomes 3 times larger. To do so simply increase each side by 7 by adding 7 to the original values.

$(x+7)(x+14)$

And multiply our area by 3.

$3(x^{2} +7x)$

set these equal to each other to find your new equation.

$(x+7)(x+14)=3(x^{2} +7x)$

Now you need to solve for $x$ . To do this first muliply $(x+7)$ and $(x+14)$ .

$(x+7)(x+14)=\\x^{2} +14x+7x+98=\\x^{2} +21x+98$

Then multiply $3(x^{2} +7x)$

$3(x^{2} +7x)=\\3x^{2} +21x$

Now you can begin solving for $x$ .

$x^{2} +21x+98=3x^{2} +21x$

Subtract $x^{2}$ from both sides.

$21x+98=2x^{2} +21x$

Subtract $21x$ from both sides.

$98=2x^{2}$

Divide by 2.

$49=x^{2}$

And finally take the sqaure root of both sides.

$\sqrt{x^{2} }=\sqrt{49}\\x=7$

Remember, because we are dealing with length, there cannot be negative. So while normally we would get both +7 <em>and</em> -7, in this case we <em>only </em>get +7.

Now that we have the value of $x$ , we can plug it into our original values.

For width we simply get 7.

For length we get 7+7=14

To check our answer we cna multiply 7 by 14 to get 98. Then we can add 7 to our length and width to get 14 and 21. Multiply these together and we get 294. 294 divided by 3 is 98, proving our answer correct.

7 0

3 years ago

Simplify the square root of -18

Levart [38]

The square root of -18 is 3i√ 2

Hope this helps

4 0

3 years ago

From a practice assignment:<br>solve the following differential equation given initial conditions

hodyreva [135]

If $y' = e^y \sin(x)$ and $y(-\pi)=0$ , separate variables in the differential equation to get

$e^{-y} \, dy = \sin(x) \, dx$

Integrate both sides:

$\displaystyle \int e^{-y} \, dy = \int \sin(x) \, dx \implies -e^{-y} = -\cos(x) + C$

Use the initial condition to solve for $C$ :

$-e^{-0} = -\cos(-\pi) + C \implies -1 = 1 + C \implies C = -2$

Then the particular solution to the initial value problem is

$-e^{-y} = -\cos(x) - 2 \implies e^{-y} = \cos(x) + 2$

(A)

4 0

1 year ago