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3 years ago

Somebody help this is algebra

Mathematics

2 answers:

sergejj [24]3 years ago

8 0

The answer is 11

You are meant to plug in 5 for x
So: h(5) = 2(5) + 1
h(5) = 11

jeyben [28]3 years ago

3 0

Answer:

2 is your answer

Step-by-step explanation:

If h(x)= 5 then 5= 2x +1

so 5 = 2x + 1

-1 -1 on both sides

then 2x = 4

2 2 divide by 2

which x = 2

Hope my answer has helped you!

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mars1129 [50]

Answer:

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4 years ago

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Anestetic [448]

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3 0

3 years ago

Once each day, Erika tracks the depth of the water in her local creek. Her first nine measurements, in inches, are: 16, 15, 13,

solniwko [45]

Answer:

Median = 13

Step-by-step explanation:

We are given the following data:

16, 15, 13, 12, 17, 14, 11, 9, 11

Formula:

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

Sorted data: 9, 11, 11, 12, 13, 14, 15, 16, 17

Median =

$=\dfrac{9+1}{2}^{th} = 5^{th}\text{ term} = 13$

Thus, the median of the given data is 13.

4 0

3 years ago

3. Complete the equivalent ways of describing the portion of the race that the

Scilla [17]

Answer:

(a). $\frac{33}{40}$ ; (b). 33 to 40

Step-by-step explanation:

Use "magic fraction" $\frac{Part}{Whole}$

(a). Distance covered by runner is 33 yards

Whole distance is 40 yards

The portion of the race that the runner has completed is

$\frac{33}{40}$

(b). The portion of the race that the runner has completed is 33 to 40

4 0

3 years ago

A rate of deposit of dust from unpolluted air is 10.0 tons per square mile per month (30 days). what is this dust fall, in milli

valina [46]

The rate of dust deposition is
$Rate= \frac{10 \,tons}{mi^{2}-30\,days} = \frac{1}{3} \, \frac{tons}{mi^{2}-day}$

Note that
1 ton = 9.07185 x 10⁵ g = 9.07185 x 10⁸ mg
1 mi = 1609.34 m = 1.60934 x 10³ m
1 day = 24 h

Therefore
$Rate=( \frac{1}{3}\, \frac{tons}{mi^{2}-day} )*( \frac{1}{24} \, \frac{day}{h} )*( \frac{1}{1.60934\times10^{3}}\, \frac{mi}{m} )^{2}*(9.07185\times10^{8}\, \frac{mg}{ton})\\= 4.865\, \frac{mg}{m^{2}-h}$

Answer: $4.865\, \frac{mg}{m^{2}-h}$

3 0

3 years ago