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Natalka [10]
3 years ago
8

87+56=____ (f r e e p o i n t s) prolly gonna give brainliest

Mathematics
1 answer:
pogonyaev3 years ago
5 0

Answer: 143

Step-by-step explanation:

Thanks for the free points

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Joan is building a sandbox in the shape of a regular pentagon. The perimeter of the pentagon is 35y4 – 65x3 inches. What is the
PIT_PIT [208]
From the choices above the answer would be: D 7y^4-13x^3 inches
8 0
3 years ago
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a person pays a loan of Rs. 3900 in monthly installments each in installment being less than the former by Rs. 20 ,the amount of
musickatia [10]

Answer:

15 installments

Step-by-step explanation:

3900 = Σ \left \ {{x=n} \atop {x=1}} \right.   400 - 20*(n - 1)

4 0
2 years ago
Help meh pwease <3 Have a great day y’all!
Alexxandr [17]

The first thing you would do is add all the numbers to get the total budget : $1500

Next you would calculate the present to get : 24.2% or just 24% if your rounding

3 0
3 years ago
A candle maker adds lemon oil to wax to make the lemon-scented candle shown. How many ounces of oil will he need for the candle
liraira [26]

Answer:

The ounces of oil needed is 5 ounces of oil

Step-by-step explanation:

The first thing to do here is to calculate the volume of the lemon-scented candle given.

Looking at shape the volume can be calculated using the formula L * B * H

where L(length) = 10cm , B(breadth) = 8cm and Height(h) = 25cm

The volume V is thus = 10 * 8 * 25 = 2,000 cm^3

The ounces of oil needed for the candle to have 0.0025 ounces of oil per cm^3 of wax will be = The volume of the lemon-scented candle * 0.0025 ounces of oil per cm^3 of wax

That will be = 0.0025 * 2,000 = 5 ounces

3 0
3 years ago
Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.
adoni [48]

Answer: The required derivative is \dfrac{8x^2+18x+9}{x(2x+3)^2}

Step-by-step explanation:

Since we have given that

y=\ln[x(2x+3)^2]

Differentiating log function w.r.t. x, we get that

\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}

Hence, the required derivative is \dfrac{8x^2+18x+9}{x(2x+3)^2}

3 0
3 years ago
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