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2 years ago

Which equation is equivalent to the given equation? - 4(x – 5) + 8x = 9x – 3

Mathematics

2 answers:

Rama09 [41]2 years ago

4 0

Answer:

Step-by-step explanation:

-4x +20 +8x = 9x -3. That's equall to 4x+20=9x-3. subtract 9x both sides, -5x+20=-3. subtract 20 both sides, -5x=-23.

seraphim [82]2 years ago

3 0

Hi!

-4(x - 5) + 8x = 9x - 3

-4x + 20 + 8x = 9x - 3

-4x + 8x - 9x = -3 - 20

Answer: B

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Graph the linear equation. Find three points that solve the equation, then plot on the graph. -3y=x-6

nata0808 [166]

Answer:

Step-by-step explanation: x - 6

The given equation can be re-written as y = ---------

-3

Arbitrarily choose x = 0. Then:

x - 6 0-6

y = --------- = ----------- = 2, so (0, 2) is a point on the graph which is also the

-3 -3 y-intercept

Arbitrarily choose x = 6. Then y = 0, and (6, 0) is another point on the graph

which happens to be the x-intercept

arbitrarily choose x = 12. Then y = (12 - 6) / (-3) = -2. Then (12, -2) is

another point on the

graph.

Plot (12, -2), (6, 0) and (0, 2). Draw a line through these three points.

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2 years ago

Complete the steps in solving the quadratic function 7x – 9 = 7x2 – 49x by completing the square. –9 = 7x2 – x –9 + = 7(x2 – 8x

anzhelika [568]

Answer:

A) 56 . . . . . . the (negative) sum of -7 and -49

b) 112 . . . . . the product of 7 and 16

c) 16 . . . . . . the square of 8/2

8 0

2 years ago

Read 2 more answers

The first-, second-, and third-year enrollment values for a technical school are shown in the table below.

VARVARA [1.3K]

Answer:

The solution to f(x) = t(x) is x = 2010

Option 3 is true.

Step-by-step explanation:

The first-year , second-year , and third-year enrollment values for a technical school are shown in the table below.

Year (x) First Year f(x) Second Year s(x) Third Year t(x)

2009 785 756 756

2010 740 785 740

2011 690 710 781

2012 732 732 710

2013 781 755 800

Now we will check each option.

Option 1: The solution to f(x) = s(x) is x = 2,009

In year 2009, f(x)=s(x)

But 785≠756

Thus, False

Option 2: The solution to f(x) = s(x) is x = 785

x represents year, but 785 it no year

Thus, False

Option 3: The solution to f(x) = t(x) is x = 2010

In year 2010, f(x)=t(x)=740

But 740=740

Thus, True

Option 4: The solution to f(x) = t(x) is x =740

x represents year, but 740 it no year

Thus, False

8 0

2 years ago

Food allergies affect an estimated 7% of children under age 5 in the US. What is the probability that in a kindergarden class of

bezimeni [28]

Answer:

Probability of atleast one of 12 student has food allergies ≈ 0.58 ( approx)

Step-by-step explanation:

Given: Probability of a children under age 5 has food allergies = 7%

= $\frac{7}{100}$

To find : Probability of atleast one of 12 student has food allergies

Probability of a chindren under age 5 does not have food allergies = $1-\frac{7}{100}$

⇒ prob = $\frac{93}{100}$

now we find Probability of atleast one of 12 student has food allergies this means we have to find prob of 1 student, 2 student, 3 student, till 12 student have allergy out of 12 student of class then add all prob.

But instead of finding all these probability we find probability of student having no allergy.i.e., 0 student then subtract it from 1(total probability)

Probability of 0 student having allergy out of 12 student = $(\frac{93}{100})^{12}$