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Gnom [1K]
4 years ago
12

7m+12 over 8= 2m-1 over 3​

Mathematics
1 answer:
Sonja [21]4 years ago
4 0

Answer:

m=-44/5

Step-by-step explanation:

(7m+12)/8=(2m-1)/3

cross product

8(2m-1)=3(7m+12)

16m-8=21m+36

16m-21m-8=36

-5m-8=36

-5m=36+8

-5m=44

m=44/-5

m=-44/5

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The market sells 1 dozen bagels
PIT_PIT [208]

Answer:

market

Step-by-step explanation:

<u>Market</u>

1 dozen bagels = 12 bagels

⇒ Cost per bagel = $7.00 ÷ 12 = $0.58 (nearest cent)

<u>Bagel shop</u>

Cost per bagel = $0.60

As  $0.58 < $0.60 the market is a better buy

3 0
2 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cint%20t%5E2%2B1%20%5C%20dt" id="TexFormula1" title="\frac{d}{dx} \
Kisachek [45]

Answer:

\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2}  t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2

Step-by-step explanation:

\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2}  t^2+1 \ \text{dt} = \ ?

We can use Part I of the Fundamental Theorem of Calculus:

  • \displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

  • \displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

  • \displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

  • \displaystyle\int\limits^b_a \text{f(t) dt}\  = -\int\limits^a_b \text{f(t) dt}

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

  • \displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx}  \int\limits^{x^2}_0 t^2+1 \text{ dt}  

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

  • \displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u]
  • where u represents any function other than a variable

For the first term, replace \text{t} with 2x, and apply the chain rule to the function. Do the same for the second term; replace

  • \displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)  

Simplify the expression by distributing 2 and 2x inside their respective parentheses.

  • [-(8x^2 +2)] + (2x^5 + 2x)
  • -8x^2 -2 + 2x^5 + 2x

Rearrange the terms to be in order from the highest degree to the lowest degree.

  • \displaystyle2x^5-8x^2+2x-2

This is the derivative of the given integral, and thus the solution to the problem.

6 0
3 years ago
Solve the inequality − 3 x + 6 &gt; 15
Leviafan [203]

Answer:

The Answer to your question is x<-7 Hope This could Help

Step-by-step explanation:

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3 years ago
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What are the zeros of the function f(x) = x2 + 5x + 5 written in simplest radical form?
RideAnS [48]

Step-by-step explanation:

The function is as follows :

f(x) = x^2 + 5x + 5

We need to find the zeroes of the function in the simplest radical form.The zero of the above function is given by :

x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}

Here,

b = 5

a = 1

c = 5

So,

x=\dfrac{-5\pm \sqrt{5^2-4\times 1\times 5} }{2(1)}\\\\x=\dfrac{-5\pm \sqrt{25-20} }{2(1)}\\\\x=\dfrac{-5\pm \sqrt{5} }{2}

Hence, the correct option is (c).

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The formula below is used to convert a temperature in degrees Celsius c to a temperature in degrees Fahrenheit f : f=1.8c+32 the
vredina [299]
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3 years ago
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