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3 years ago

Please show work for both problems

Mathematics

1 answer:

pentagon [3]3 years ago

5 0

Answer:

1) x = 25 2) x = 23; m∠CBD = 80 degrees

Step-by-step explanation:

1) 5x + 15 = 6x - 10 Set both equations equal to each other

- 5x - 5x Subtract 5x from both sides

15 = x - 10

+ 10 + 10 Add 10 to both sides

25 = x

2) 2x + 14 + x + 7 = 90 Set the equations equal to 90

3x + 21 = 90 Combine like terms

- 21 - 21 Subtract 21 on both sides

3x = 69 Divide both sides by 3

x = 23

Plug 23 into the equation

2(23) + 14 = 60

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Is my answer correct? If not what’s the right answer

pochemuha

Answer:

Yup, 27 is correct!

Step-by-step explanation:

5(7)-8

= 35 - 8

= 27

8 0

3 years ago

Read 2 more answers

James wants to promote his band on the internet. Site A offers website hosting for $4.95 per month with a $49.95 startup fee. Si

cupoosta [38]

Answer:

Part 1) see the procedure

Part 2) $4.95x+49.95 < 9.95 x$

Part 3) $x > 9.99\ months$

Part 4) The minimum number of months, that he needs to keep the website for site A to be less expensive than site B is 10 months

Step-by-step explanation:

Part 1) Define a variable for the situation.

Let

x ------> the number of months

y ----> the total cost monthly for website hosting

Part 2) Write an inequality that represents the situation.

we know that

Site A

$y=4.95x+49.95$

Site B

$y=9.95x$

The inequality that represent this situation is

$4.95x+49.95 < 9.95 x$

Part 3) Solve the inequality to find out how many months he needs to keep the website for Site A to be less expensive than Site B

$4.95x+49.95 < 9.95 x$

Subtract 4.95x both sides

$4.95x+49.95-4.95x < 9.95 x-4.95x$

$49.95 < 5x$

Divide by 5 both sides

$49.95/5 < 5x/5$

$9.99 < x$

Rewrite

$x > 9.99\ months$

Part 4) describe how many months he needs to keep the website for Site A to be less expensive than Site B.

The minimum number of months, that he needs to keep the website for site A to be less expensive than site B is 10 months

4 0

3 years ago

PLEASE HELP ASAP! I don’t recall how to do this!

MakcuM [25]

Answer:

Step-by-step explanation:

For a. we start by dividing both sides by 200:

$(1.05)^x=1.885$

In order to solve for x, we have to get it out from its position of an exponent. Do that by taking the natural log of both sides:

$ln(1.05)^x=ln(1.885)$

Applying the power rule for logs lets us now bring down the x in front of the ln:

x * ln(1.05) = ln(1.885)

Now we can divide both sides by ln(1.05) to solve for x:

$x=\frac{ln(1.885)}{ln(1.05)}$

Do this on your calculator to find that

x = 12.99294297

For b. we will first apply the rule for "undoing" the addition of logs by multipllying:

$ln(x*x^2)=5$

Simplifying gives you

$ln(x^3)=5$

Applying the power rule allows us to bring down the 3 in front of the ln:

3 * ln(x) = 5

Now we can divide both sides by 3 to get

$ln(x)=\frac{5}{3}$

Take the inverse ln by raising each side to e:

$e^{ln(x)}=e^{\frac{5}{3}}$

The "e" and the ln on the left undo each other, leaving you with just x; and raising e to the power or 5/3 gives you that

x = 5.29449005

For c. begin by dividing both sides by 20 to get:

$\frac{1}{2}=e^{.1x}$

"Undo" that e by taking the ln of both sides:

$ln(.5)=ln(e^{.1x})$

When the ln and the e undo each other on the right you're left with just .1x; on the left we have, from our calculators:

-.6931471806 = .1x

x = -6.931471806

Question d. is a bit more complicated than the others. Begin by turning the base of 4 into a base of 2 so they are "like" in a sense:

$(2^2)^x-6(2)^x=-8$

Now we will bring over the -8 by adding:

$(2^2)^x-6(2)^x+8=0$

We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution. Let's let $u=2^x$

When we do that, we can rewrite the polynomial as

$u^2-6u+8=0$

This factors very nicely into u = 4 and u = 2

But don't forget the substitution that we made earlier to make this easy to factor. Now we have to put it back in:

$2^x=4,2^x=2$

For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:

$2^2=2^x$

Now that the bases are the same, we can say that

x = 2

For the second solution, we will raise the 2 on the right to a power of 1 to get:

$2^x=2^1$

Now that the bases are the same, we can say that

x = 1

5 0

3 years ago

35xy/7y simplified plz just simple fied it ;-; rip me i hate math mevr udnerstood it

ch4aika [34]

Its 7xy/1y I understand it its pretty easy

5 0

3 years ago

WHAT IS THE SQUARE ROOT OF POTATO?

SIZIF [17.4K]

Answer:

Do you mean square root of pi?

Step-by-step explanation:

Because if it is then the answer is 1.772453850905516

8 0

3 years ago