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4 years ago

8

3x-23x+20

3x {2} - 23 + 20" align="absmiddle" class="latex-formula">

1 answer:

True [87]4 years ago

7 0

I'm pretty sure it's -20X-23+20etx

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An object is launched at 64 ft/sex

Zolol [24]

64ft/sex lol per sex?

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3 years ago

Write 2 ways to take apart 5.

fenix001 [56]

4 and 1
3 and 2
im pretty sure thats is the answer, the question is quite vague.

8 0

3 years ago

SOlve for x+2 = 2x-9+4

Oksi-84 [34.3K]

Answer:

x = 7

Step-by-step explanation:

Step 1. Combine like terms for simplification

x + 2 = 2x - 5

Step 2. Subtact 2 from both sides

x = 2x - 7

Step 3. Subtract 2x from both sides

-x = -7

Step 4. Divide it all by -1 to isolate the variable.

x = 7

5 0

3 years ago

52.Consider the functionsf(x) = 2* x+6 and g(x) = 5*2x.

likoan [24]

Answer:

$x=\frac{6\ln \left(2\right)}{2\ln \left(5\right)-\ln \left(2\right)}\\x \approx 1.65$

Step-by-step explanation:

Given

$f(x) = 2^{\left(x+6\right)}$

$g(x) = 5^{2x}$

We want to find $f(x) = g(x)$

For this you need to:

if $f(x) = g(x)$ , then $\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)$

$\ln \left(2^{x+6}\right)=\ln \left(5^{2x}\right)$

Apply log rule: $\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$\left(x+6\right)\ln \left(2\right)=2x\ln \left(5\right)$

Solve for x

Expand $\left(x+6\right)\ln \left(2\right) = \ln \left(2\right)x+6\ln \left(2\right)$

$\ln \left(2\right)x+6\ln \left(2\right)=2x\ln \left(5\right)$

$\ln \left(2\right)x=2x\ln \left(5\right)-6\ln \left(2\right)\\\ln \left(2\right)x-2x\ln \left(5\right)=-6\ln \left(2\right)\\\left(\ln \left(2\right)-2\ln \left(5\right)\right)x=-6\ln \left(2\right)\\\\x=\frac{6\ln \left(2\right)}{2\ln \left(5\right)-\ln \left(2\right)}\approx 1.65$

8 0

3 years ago

A garrison of 400men had food for 40days.After 10 days,200 more men joined them .How long will the food last now?(Assume that th

mestny [16]

Let $x$ be the amount of food needed for one man for one day. At the beginning, we have enough food for 400 men for 40 days, i.e.

$400\cdot 40\cdot x = 16000x$

After 10 days, the 400 men will have consumed

$400\cdot 10\cdot x = 4000x$

food, impliying that the food remaining is

$16000x-4000x=12000x$

If 200 more men join the garrison, there are now 600 men. Each of them requires $x$ food each day, and there are $12000x$ units of food remaining. They will last

$\dfrac{12000x}{600}=20$ days.

6 0

4 years ago