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alexandr402 [8]

2 years ago

15

Solve for w: w+6tw=11-w

1 answer:

Dmitrij [34]2 years ago

7 0

<span>w+6tw=11-w
</span>
<span>Let's solve for w.
</span><span><span>
w+<span><span>6t</span>w</span></span>=<span>11−w

</span></span>
Step 1: Add w to both sides.<span><span><span><span><span>
6t</span>w</span>+w</span>+w</span>=<span><span><span>−w</span>+11</span>+w</span></span><span><span><span><span>
6t</span>w</span>+<span>2w</span></span>=11</span><span>
Step 2: Factor out variable w.
</span><span><span>
w<span>(<span><span>6t</span>+2</span>)</span></span>=11</span><span>
Step 3: Divide both sides by 6t+2.
</span><span><span>
w<span>(<span><span>6t</span>+2</span>)</span></span><span><span>/6t</span>+2</span></span>=<span>11/<span><span>6t</span>+2</span></span>

Answer: 11/6t+2

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2 years ago

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3 years ago

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Find the maximum value or minimum value for the function f(x) = 0.15(x + 1)² - 3.

il63 [147K]

Answer:

The minimum value for $f(x) = 0.15(x + 1)^2 - 3$ is $-3$ .

Step-by-step explanation:

Given function is $f(x) = 0.15(x + 1)^2 - 3$

We need to find the maximum value or the minimum value for the function.

Now, differentiate $f(x) = 0.15(x + 1)^2 - 3$ w.r.t $x$ .

$f'(x) =\frac{d}{dx}(0.15(x + 1)^2 - 3)\\f'(x)=\frac{d}{dx}(0.15(x+1)^2-\frac{d}{dx}(3)\\$

$f'(x)=2\times 0.15(x+1)\frac{d}{dx}(x+1)-0\\f'(x)=0.3(x+1)(1)\\f'(x)=0.3(x+1)$

Now, we will equate $f'(x)=0$ to find critical point.

$0.3(x+1)=0\\x=-1$

Plug this critical point in to the function $f(x) = 0.15(x + 1)^2 - 3$ we get,

$f(-1) = 0.15(-1 + 1)^2 - 3\\f(-1)=-3$

Also, $f''(x)=0.3$ which is positive, We have minimum value.

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3 years ago

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Answer:

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To solve this equation, plug your x value in and simplify.

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3 0

2 years ago

Read 2 more answers

If EF = 4x-15, FG=3x-7, and EG =20, find the value of x, EF and FG. The drawing is not to scale.

Andrej [43]

Step-by-step explanation:

EF = 4x - 15

FG = 3x - 7

EG = EF + FG = 20

so,

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2 years ago