All categories

3 years ago

Please help thank you very much. 20 points.

Mathematics

2 answers:

Goryan [66]3 years ago

8 0

The correct answer is S.

qaws [65]3 years ago

7 0

Answer is the letter S

You might be interested in

Mr roosevelt has 48 nails that each weigh 1.35 onnces what is the weight of these nails in onces

Harrizon [31]

Answer:

64.8

Step-by-step explanation:

maybe try multiplying the ounces by the number of nails

7 0

3 years ago

you bought your friends a heart shaped pound of chocolate caramels. You anticipate each chocolate was 50 calories. Each chocolat

Wittaler [7]

There was 25.37% error in calculations.

Step-by-step explanation:

Given,

Anticipated calories = Approx value = 50 calories

Actual calories = Exact value = 67 calories

Percent error = $\frac{|aprrox\ value-exact\ value|}{exact\ value}*100$

Percent error = $\frac{|50-67|}{67}*100$

Percent error = $\frac{|-17|}{67}*100=\frac{17}{67}*100$

Percent error = $\frac{1700}{67}$

Percent error = 25.37%

There was 25.37% error in calculations.

Keywords: percent error, division

Learn more about division at:

#LearnwithBrainly

8 0

3 years ago

PLEASE HELP. TAKE AT HOME TEST. if u = LN5 and v = LN2, write LN8/15 in terms of u and v. (8/15 is 8 over 25)

astra-53 [7]

Answer:

$\ln( \frac{8}{25} ) = 3v - 2u$

Step-by-step explanation:

We have that

$u= \ln(5)$

and

$v = \ln2$

This implies that;

$5 = {e}^{u}$

and

$2 = {e}^{v}$

Now find and expresion for

$\ln( \frac{8}{25} )$

Which is

$\ln( \frac{8}{25} ) = ln(8) - ln(25)$

$\ln( \frac{8}{25} ) = ln( {2}^{3} ) - ln( {5}^{2} )$

$\ln( \frac{8}{25} ) = 3 ln( {2} ) - 2ln(5)$

$\ln( \frac{8}{25} ) = 3v - 2u$

8 0

3 years ago

Helpppppopoopppppppppppp

VARVARA [1.3K]

Tonis donation is 3 times longer (6 x 3 = 18)

8 0

3 years ago

Read 2 more answers

Find the domain of <br> f(x) = - squareroot x^2 - 4

erma4kov [3.2K]

Answer:

d. x ≥ 2 or x ≤ -2

Step-by-step explanation:

The domain is the set of all possible x-values that will make the function work.

For the function to work, the radicand cannot be negative.

x² - 4 ≥ 0

x² ≥ 4

x ≥ 2 -x ≥ 2

x ≤ -2

The domain is x ≥ 2 and x ≤ -2.

The graph of your function shows that ƒ(x) does not exist when -2 < x < 2.

6 0

3 years ago