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3 years ago

In the library at Lenape Elementary School, there are 3/8

Mathematics

2 answers:

antoniya [11.8K]3 years ago

6 0

Your answer is 163⁄<span>200 i thinks because if this is adding yeah</span>

goblinko [34]3 years ago

6 0

It's 44 divided by 3/8. Whatever the solution is to that. That will be your correct anwser

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Item 4

KonstantinChe [14]

Answer:

oscar raises 246 dollars.

Step-by-step explanation:

you add 23.25 to amys amount and then times that by 3 to find the answer.

6 0

3 years ago

Read 2 more answers

Use the descriminant to describe the roots of each equation. Then select the best description. 16x^2+8x+1=0

Pachacha [2.7K]

Use the values of a, b, and c to find the discriminant:
0
I don't know if this is correct

8 0

3 years ago

Need asap!!!!!!!!!!!

katrin2010 [14]

Answer:

50 J

Step-by-step explanation:

According to energy conservative law energy can not be destroyed. So as the potential energy has been decreased to 50 J, the decreased amount has become kinetic energy.

100 - 50 = 50

3 0

2 years ago

The area of a square garden is 98 m2. How long is the diagonal?

stiks02 [169]

Answer:

The diagonal is about 14 m

Step-by-step explanation:

A = s^2

98 = s^2

s = √98

diagonal c^2 = 98 + 98

c^2 = 196

c = √196

c ≈ 14

Answer:

The diagonal is about 14 m

8 0

3 years ago

Read 2 more answers

Help ASAP show work please thanksss!!!!

Llana [10]

Answer:

$\displaystyle log_\frac{1}{2}(64)=-6$

Step-by-step explanation:

<u>Properties of Logarithms</u>

We'll recall below the basic properties of logarithms:

$log_b(1) = 0$

Logarithm of the base:

$log_b(b) = 1$

Product rule:

$log_b(xy) = log_b(x) + log_b(y)$

Division rule:

$\displaystyle log_b(\frac{x}{y}) = log_b(x) - log_b(y)$

Power rule:

$log_b(x^n) = n\cdot log_b(x)$

Change of base:

$\displaystyle log_b(x) = \frac{ log_a(x)}{log_a(b)}$

Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

$\displaystyle log_\frac{1}{2}(64)$

Factoring $64=2^6$ .

$\displaystyle log_\frac{1}{2}(64)=\displaystyle log_\frac{1}{2}(2^6)$

Applying the power rule:

$\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}(2)$

Since

$\displaystyle 2=(1/2)^{-1}$

$\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}((1/2)^{-1})$

Applying the power rule:

$\displaystyle log_\frac{1}{2}(64)=-6\cdot log_\frac{1}{2}(\frac{1}{2})$

Applying the logarithm of the base:

$\mathbf{\displaystyle log_\frac{1}{2}(64)=-6}$

5 0

3 years ago