E-1/2f when e=15 and f=2

Hey!<br> Question: <br> Simplify (16y^8)^3/4<br><br> Thanks for the help!

olasank [31]

Hope this helps! Is from the app Photomath.

8 0

3 years ago

If f (x) = StartRoot one-half x minus 10 EndRoot + 3, which inequality can be used to find the domain of f(x)?

just olya [345]

The inequality that represents the domain of the function is $\frac 12x - 10 \ge 0$

<h3>How to determine the domain?</h3>

The function is given as:

$f(x) = \sqrt{\frac 12x - 10} + 3$

Set the radicand greater than or equal to 0

$\frac 12x - 10 \ge 0$

Multiply through by 2

x - 20 ≥ 0

Add 20 to both sides

x ≥ 20

Hence, the domain of the function is x ≥ 20

Read more about domain at:

brainly.com/question/1770447

#SPJ1

7 0

1 year ago

Read 2 more answers

Whats 4485 square feet of grass in 3/4 of an hour. How many square feet of grass can be fertilized per hour

alex41 [277]

4485 square feet of grass ... 3/4 of an hour
x square feet of grass = ? ... 1 hour

If you would like to know how many square feet of grass can be fertilized per hour, you can calculate this using the following steps:

4485 * 1 = 3/4 * x
4485 = 3/4 * x /*4/3
x = 4485 * 4 / 3
x = 5980 square feet

Result: 5980 square feet of grass can be fertilized in one hour.

7 0

2 years ago

⚠URGENT!!!⚠ <br><br> solve for m.

Zinaida [17]

Answer:

Step-by-step explanation:

Remark

Read the following carefully.

There is a beautiful theorem that has to do with the endpoints of two angles sharing the same endpoints.

To be a little clearer, I hope, that makes < BAC = <BDC because both angles have B and C as their endpoints inside the circle. Make sure you understand that statement before moving on.

For this problem <BDC = <CAB = 33 degrees.

That means that ADC = 37 + 33 = 70

Solution

<ADC and CBA are opposite angles.

That means that they add to 180

From the above statement in the Remark section <ADC = 37 + 33 = 70 degrees <ABD + <DBC = <ABC = m + 71

<ABC + ADC = 180

m + 71 + 70 = 180 Combine

m + 141 = 180 Subtract 141 from both sides.

m+141-141= 180 - 141 Combine

m = 39

Answer: m = 39

3 0

1 year ago

A jeweler wants to make 14 grams of an alloy that is precisely 75% gold.. The jeweler has alloys that are 25% gold, 50% gold, &a

Goryan [66]

Given that the jeweler has alloys that are 25% gold, 50% gold, and 82% gold.

As he wants to make 14 grams of an alloy by adding two different alloys that is precisely 75% gold, so one alloy must have a percentage of gold more than 75%.

One alloy is 82% gold and, the second can be chosen between 25% gold, 50% gold, so there are two cases.

Case 1: 82% gold + 50% gold

Let x grams of 82% gold and y grams of 50% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 50% of y

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times y \\\\$

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times (14-x)$ [as x+y=14]

$\Rightarrow 75 \times 14 = 82 \times x + 50 \times (14-x) \\\\\Rightarrow 75 \times 14 = 82 \times x + 50 \times14-50\times x \\\\\Rightarrow 75 \times 14 = 32 \times x + 50 \times14 \\\\\Rightarrow 32 \times x =75 \times 14 - 50 \times14 \\\\$

$\Rightarrow x =(25 \times 14)/32=10.9375$ grams

and $y = 14-x= 14-10.9375=3.0625$ grams.

Hence, 10.9375 grams of 82% gold and 3.0625 grams of 50% gold added to make 14 grams of 75% gold.

Case 2: 82% gold + 25% gold

Let x grams of 82% gold and y grams of 25% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 25% of y

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times y \\\\\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times (14-x) \\\\ \Rightarrow 75 \times 14 = 82 \times x + 25 \times (14-x) \\\\\Rightarrow 75 \times 14 = 82 \times x + 25 \times14-25\times x \\\\\Rightarrow 75 \times 14 = 57 \times x + 25 \times14 \\\\\Rightarrow 57 \times x =75 \times 14 - 25 \times14 \\\\$

$\Rightarrow x =(50 \times 14)/57=12.28$ grams

and $y = 14-x= 14-12.28=1.72$ grams.

Hence, 12.28 grams of 82% gold and 1.72 grams of 50% gold added to make 14 grams of 75% gold.

3 0

2 years ago