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Vikentia [17]
3 years ago
13

Find the Area of a square with length(x + 7) and width (x + 7)

Mathematics
1 answer:
ValentinkaMS [17]3 years ago
8 0

Answer:

x^2 +14x+49

Step-by-step explanation:

Area = length * width

        = ( x+7) * ( x+7)

       FOIL

    first: x^2

    outer: 7x

   inner: 7x

    last: 7*7 = 49

Add them together

x^2 +7x+7x + 49

Combine like terms

x^2 +14x+49

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12x+5=4x? Show steps, please.
gavmur [86]

Answer:

x = 5/-8

Step-by-step explanation:

4x = 12x + 5

4x - 12x = 5

-8x = 5

x = 5/-8

6 0
3 years ago
Demerol 45mg and atropine 400mcg/ml give how much per ml to total volume to inject demerol contains 50mg/ml, atropine contains 4
OverLord2011 [107]

Answer:

Volume injected  = 1.65 ml

Step-by-step explanation:

mass of Demerol =45 mg.

density of pre-filled in syringe = 50 mg/ml

Volume = 45/50 = 0.9 ml

For atropine mass = 0.3 mg

density= 400 mcg/ml     [ Note : 1 mcg = 0.001 mg]

volume filled  = 0.3/400(0.001) = 0.75 ml

So, the total volume filled = 0.75+0.9 = 1.65 ml

6 0
3 years ago
Help me please!!! I will give Brainly if correct!
Alexxandr [17]

Answer:

B, 20mph

Step-by-step explanation:

540 miles in 27 hours is 540/27 = 20 miles in 1 hour. This is 20 mph

3 0
3 years ago
Read 2 more answers
What is the equation in standard form of the line that passes through the point (1, 24) and has a slope of -0.6.
lana [24]

Answer:

0.6x+y=24.6

Step-by-step explanation:

y-y1=m(x-x1)

y-24=-0.6(x-1)

y=-0.6x+0.6+24

y=-0.6x+24.6

y-(-0.6x)=24.6

y+0.6x=24.6

0.6x+y=24.6

6 0
3 years ago
A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies
RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

                              Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

8 0
3 years ago
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