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2 years ago

What is the solution to this equation? log_7 (5 + 11x) = 2

Mathematics

1 answer:

11111nata11111 [884]2 years ago

5 0

Answer:

x = 4

Step-by-step explanation:

Using the rule of logarithms

$log_{b}$ x = n , then x = $b^{n}$

Given

$log_{7}$ (5 + 11x) = 2 , then

5 + 11x = 7² = 49 ( subtract 5 from both sides )

11x = 44 ( divide both sides by 11 )

x = 4

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Nicole sold 1 cherry pie and 9 pumpkin pies for $60. Lisa sold 11 cheery pies and 4 pumpkin pies for $90. what's the cost each o

zubka84 [21]

X= cost per cherry pie
y= cost per pumpkin pie

NICOLE
1x + 9y= $60

LISA
11x + 4y= $90

STEP 1
multiply Nicole's equation by -11

-11(1x + 9y)= -11($60)
multiply -11 by all terms

(-11 * x) + (-11 * 9y)= (-11 * 60)

-11x - 99y= -660

STEP 2
add Nicole's new equation from step 1 to Lisa's equation to solve for y (using the elimination method)

-11x - 99y= -660
11x + 4y= 90
the x terms "cancel out"

-95y= -570
divide both sides by -95

y= $6 per pumpkin pie

STEP 3
substitute y value into either original equation to solve for x

x + 9y= $60

x + 9(6)= 60

x + 54= 60
subtract 54 from both sides

x= $6 per cherry pie

CHECK
11x + 4y= $90
11(6) + 4(6)= 90
66 + 24= 90
90= 90

ANSWER: Each cherry pie costs $6 and each pumpkin pie costs $6.

Hope this helps! :)

3 0

3 years ago

Can you help me pls, this doesn't make any sense

Gala2k [10]

Area of the larger square = 12² = 144 cm²
Area of the smaller square = 7² = 49 cm²
Area of the section = 144-49 = 95 cm²

7 0

3 years ago

What formula is the same as B = atra?

horrorfan [7]

Answer:

atra=B

Step-by-step explanation:

that's the only one I can think of lol

3 0

3 years ago

Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we

Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N( $\mu = 7 lbs , \sigma^{2} = 0.875^{2} lbs$ )

The standard normal z distribution is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, X bar = sample mean weight

n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{7.5-7}{\frac{0.875}{\sqrt{4} } }$ ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

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3 years ago

What is the slope of the line through (–4, 3) and (5, 3)?

Sloan [31]

Answer:

Step-by-step explanation:

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3 years ago

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