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ivanzaharov [21]
4 years ago
7

HELP PLEASE THANKS 1 more after this

Mathematics
2 answers:
trasher [3.6K]4 years ago
8 0
To find the answer, you would multiply the exponents and leave the base the same. 

Your answer would be D. 5^50

I hope this helps!
bonufazy [111]4 years ago
6 0
5=5^15 is the Answer, you add the exponent not multiply them
5^10*5^5= 30517578125
5^15=30517578125
Check your calculator
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I need help with this ^
Inga [223]

Answer:

23

Step-by-step explanation:

17 pluse 2 is 19

180 minus 19 is 161

add all the vs you get 7

so 161 divided by 7 is 23

6 0
3 years ago
Read 2 more answers
Office supplies had a normal starting balance of $75. there were debit postings of $80 and credit postings of $60 during the mon
Katena32 [7]

Answer:

I think it’s 95 debit but it could be credit

Step-by-step explanation:

75 + 80 = 155

155 - 60 = 95

5 0
3 years ago
Solve for X. <br><br> <img src="https://tex.z-dn.net/?f=log_%7B8%7D%20x-log_%7B8%7D%20%28-2x%2B4%29%20%3D%20log_%7B8%7D%203" id=
taurus [48]

Answer: x=\frac{12}{7}

Step-by-step explanation:

Remember the logarithms properties:

log(m)-log(n)=log(\frac{m}{n})\\\\b^{log_b(a)}=a

Then,simplifying:

log_{8}(\frac{x}{-2x+4}) = log_{8}(3)

Apply base 8 to boths sides and then solve for "x":

8^{log_{8}(\frac{x}{-2x+4})}=8^{log_{8}(3)}\\\\\frac{x}{-2x+4}=3\\\\x=3(-2x+4)\\\\x=-6x+12\\x+6x=12\\7x=12\\\\x=\frac{12}{7}

4 0
3 years ago
Read 2 more answers
Find the lateral area of the square pyramid shown to the nearest whole number.
Xelga [282]

Answer:

4,300

Step-by-step explanation:

Lateral area of a squared Pyramid is given as ½ × Perimeter of base (P) × slant height of pyramid

Thus, we are given,

Side base length (s) = 43 yd

height (h) = 25 yd

Let's find the perimeter

Permimeter = 4(s) = 4(43) = 172 yd

Calculate the slant height using Pythagorean theorem.

Thus, l² = s²+h²

l² = 43²+25² = 1,849+625

l² = 2,474

l = √2,474

l ≈ 50 yd

=>Lateral area = ½ × 172 × 50

= 172 × 25

= 4,300 yd

3 0
3 years ago
Assuming that the equation defines x and y implicitly as differentiable functions xequals​f(t), yequals​g(t), find the slope of
Doss [256]

Answer:

\dfrac{dx}{dt} = -8,\dfrac{dy}{dt} = 1/8\\

Hence, the slope , \dfrac{dy}{dx} = \dfrac{-1}{64}

Step-by-step explanation:

We need to find the slope, i.e. \dfrac{dy}{dx}.

and all the functions are in terms of t.

So this looks like a job for the 'chain rule', we can write:

\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx} -Eq(A)

Given the functions

x = f(t)\\y = g(t)\\

and

x^3 +4t^2 = 37 -Eq(B)\\2y^3 - 2t^2 = 110 - Eq(C)

we can differentiate them both w.r.t to t

first we'll derivate Eq(B) to find dx/dt

x^3 +4t^2 = 37\\3x^2\frac{dx}{dt} + 8t = 0\\\dfrac{dx}{dt} = \dfrac{-8t}{3x^2}\\

we can also rearrange Eq(B) to find x in terms of t , x = (37 - 4t^2)^{1/3}. This is done so that \frac{dx}{dt} is only in terms of t.

\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\\dfrac{dx}{dt} = \dfrac{-8(3)}{3(37 - 4(3)^2)^{2/3}}\\\dfrac{dx}{dt} = -8

now let's differentiate Eq(C) to find dy/dt

2y^3 - 2t^2 = 110\\6y^2\frac{dy}{dt} -4t = 0\\\dfrac{dy}{dt} = \dfrac{4t}{6y^2}

rearrange Eq(C), to find y in terms of t, that is y = \left(\dfrac{110 + 2t^2}{2}\right)^{1/3}. This is done so that we can replace y in \frac{dy}{dt} to make only in terms of t

\dfrac{dy}{dt} = \dfrac{4t}{6y^2}\\\dfrac{dy}{dt}=\dfrac{4t}{6\left(\dfrac{110 + 2t^2}{2}\right)^{2/3}}\\

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

\dfrac{dy}{dt} = \dfrac{4(3)}{6\left(\dfrac{110 + 2(3)^2}{2}\right)^{2/3}}\\\dfrac{dy}{dt} = \dfrac{1}{8}

Finally we can plug all of our values in Eq(A)

but remember when plugging in the values that \frac{dy}{dt} is being multiplied with \frac{dt}{dx} and NOT \frac{dx}{dt}, so we have to use the reciprocal!

\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx}\\\dfrac{dy}{dx} = \dfrac{1}{8}.\dfrac{1}{-8} \\\dfrac{dy}{dx} = \dfrac{-1}{64}

our slope is equal to \dfrac{-1}{64}

7 0
3 years ago
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