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Andrej [43]
3 years ago
10

What is the highest common factor of 45 and 315?

Mathematics
2 answers:
7nadin3 [17]3 years ago
8 0
The highest common factor is 45
tiny-mole [99]3 years ago
3 0

The GCF of 45 and 315 is 45. So before conformation of my answer refer to your math teacher .

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SOMEONE PLZZZZ HELP THIS IS DUE SOON!!! :)
Lelu [443]

Answer:

im stumped.

Step-by-step explanation:

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3 years ago
What is the value of (-7+3i)+(2-6i)? A)-9+9i B)-5+9i C) -9-3i D) -5-3i
Delicious77 [7]

Answer:

D

Step-by-step explanation:

Add real numbers and add imaginary numbers

(-7 + 3i)+(2 - 6i) = -7 + 2 + 3i - 6i

                        =  - 5 - 3i

4 0
3 years ago
Read 2 more answers
3 over 7 = − 1 over 7 q
Finger [1]
Q= - 1/3 is the answer-move the negative sign to the left multiply both sides by 7q then cancel 7 then finally divide both sides by 3. Hope this helps
6 0
3 years ago
Write an equation of the line passing through point P(−2, 6) that is parallel to the line x=−5.
andriy [413]

Answer:

x = -2

Step-by-step explanation:

The line x = - 5 means it is a straight vertical line that has no slope. This means that the slope is undefined.

Now, for this line to be parallel to the line passing through point P(−2, 6), it means the line will be a vertical line too with the equation of form (x = a).

Thus, the equation of the line is x = -2

7 0
3 years ago
HELP ME PLEASE
solmaris [256]

Answer:

The initial mass of the sample was 16 mg.

The mass after 5 weeks will be about 0.0372 mg.

Step-by-step explanation:

We can write an exponential function to model the situation.

Let the initial amount be A. The standard exponential function is given by:

P(t)=A(r)^t

Where r is the rate of growth/decay.

Since the half-life of Palladium-100 is four days, r = 1/2. We will also substitute t/4 for t to to represent one cycle every four days. Therefore:

\displaystyle P(t)=A\Big(\frac{1}{2}\Big)^{t/4}

After 12 days, a sample of Palladium-100 has been reduced to a mass of two milligrams.

Therefore, when x = 12, P(x) = 2. By substitution:

\displaystyle 2=A\Big(\frac{1}{2}\Big)^{12/4}

Solve for A. Simplify:

\displaystyle 2=A\Big(\frac{1}{2}\Big)^3

Simplify:

\displaystyle 2=A\Big(\frac{1}{8}\Big)

Thus, the initial mass of the sample was:

A=16\text{ mg}

5 weeks is equivalent to 35 days. Therefore, we can find P(35):

\displaystyle P(35)=16\Big(\frac{1}{2}\Big)^{35/4}\approx0.0372\text{ mg}

About 0.0372 mg will be left of the original 16 mg sample after 5 weeks.

5 0
3 years ago
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