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2 years ago

Are the two expressions equivalent when x=5

Mathematics

2 answers:

adell [148]2 years ago

7 0

Answer:

Yes

Step-by-step explanation:

The first one is 35+15 because 7*5 is 35 and 3*5 is 15

The second one is 45+5 because 9*5=45

35+15=50

45+5=50

Yes, they are equal.

SpyIntel [72]2 years ago

5 0

Answer:Yes!

Step-by-step explanation:If you put 5 in every spot where the x varible is and solve the equations they both equal the answer.

7x+3x

7(5)+3(5)= 7 x 5=35, 3 x 5=15

35+15=50

9x+5

9(5)+5= 9 x 5= 45

45+5=50

Hope this helps!

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The center of a circle is at (0, 0), and the radius of the circle is 3(sqrt)2

Assoli18 [71]

Answer:

Step-by-step explanation:

The equation of a circle is denoted by: $(x-h)^2+(y-k)^2=r^2$ , where (h, k) is the center and r is the radius.

Here, we see that the center is (0, 0), so h = 0 and k = 0. Also, the radius is $3\sqrt{2}$ , so r = $3\sqrt{2}$ and $r^2=(3\sqrt{2})^2=9*2=18$ .

Then we can write our equation:

$(x-0)^2+(y-0)^2=18$

$x^2+y^2=18$

The answer is B.

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2 years ago

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AnnyKZ [126]

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Step-by-step explanation:

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2 years ago

Which kind of function best models the data in the table? Graph the data and write an equation to model the data. x y 0 –2 1 –3

jeyben [28]

Let

A (0,-2) B (1,-3) C (2,-4) D ( 3,-5) E (4,-6)

using a graph tool

see the attached figure N 1

case a) exponential

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see the attached figure N 2

case b) quadratic

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see the attached figure N 3

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see the attached figure N 4

case d) linear

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see the attached figure N 5

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2 years ago

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2 years ago

Read 2 more answers

a random sample of 4 claims are selected from a lot of 12 that has 3 nonconforming units. using the hypergeometric distribution

Sloan [31]

Answer:

The probability that the sample will contain exactly 0 nonconforming units is P=0.25.

The probability that the sample will contain exactly 1 nonconforming units is P=0.51.

Step-by-step explanation:

We have a sample of size n=4, taken out of a lot of N=12 units, where K=3 are non-conforming units.

We can write the probability mass function as:

$P(x=k)=\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$

where k is the number of non-conforming units on the sample of n=4.

We can calculate the probability of getting no non-conforming units (k=0) as:

$P(x=0)=\frac{\binom{3}{0}\binom{9}{4}}{\binom{12}{4}}=\frac{1*126}{495}=\frac{126}{495} = 0.25$

We can calculate the probability of getting one non-conforming units (k=1) as:

$P(x=1)=\frac{\binom{3}{1}\binom{9}{3}}{\binom{12}{4}}=\frac{3*84}{495}=\frac{252}{495} = 0.51$

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3 years ago