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sp2606 [1]
3 years ago
14

What is this? q+2/5 = 2q - 11/ 7 Please help.

Mathematics
1 answer:
just olya [345]3 years ago
4 0
<span>We simplify the equation to the form, which is simple to understand
<span>q+2/5=2q-11/7

Simplifying:
<span>q + 0.4=2q-11/7

Simplifying:
<span>q + 0.4=2q-1.57142857143

We move all terms containing q to the left and all other terms to the right.
<span>+1q-2q=-1.57142857143-0.4

We simplify left and right side of the equation.
<span>-1q=-1.97142857143

We divide both sides of the equation by -1 to get q.
<span>q=1.97142857143</span></span></span></span></span></span></span>
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B is the answer.

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3 years ago
A triangle has a 60° angle, and the two adjacent sides are 12 and "12 times the square root of 3". Find the radius of a circle w
shusha [124]

The radius of a circle with the same vertex as a center is 12 units

<h3>Application of Pythagoras theorem;</h3>

To get the radius of the circle, we need to determine the diameter of the circle first:

According to SOH CAH TOA:

sin\theta = \frac{opp}{hyp} \\sin60 = \frac{12\sqrt{3}}{hyp} \\hyp =\frac{12\sqrt{3}}{sin60} \\hyp =\frac{2\times12\sqrt{3}}{\sqrt3}} \\hyp = 24 = diameter

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3 years ago
Consider the equation below.
Korvikt [17]

Answer:

Equation in square form:

y=3(x+5)^2-4

Extreme value:

(h,k)=(-5,-4)

Step-by-step explanation:

We are given

y=3x^2+30x+71

we can complete square

y=3(x^2+10x)+71

we can use formula

a^2+2ab+b^2=(a+b)^2

y=3(x^2+2\times x\times 5)+71

now, we can add and subtract 5^2

y=3(x^2+2\times x\times 5+5^2-5^2)+71

y=3(x^2+2\times x\times 5+5^2)-3\times 5^2+71

y=3(x+5)^2-75+71

So, we get equation as

y=3(x+5)^2-4

Extreme values:

we know that this parabola

and vertex of parabola always at extreme values

so, we can compare it with

y=a(x-h)^2+k

where

vertex=(h,k)

now, we can compare and find h and k

y=3(x+5)^2-4

we get

h=-5

k=-4

so, extreme value of this equation is

(h,k)=(-5,-4)

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