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2 years ago

Factor 20x^2+15x-35 you must show your work

Mathematics

1 answer:

algol [13]2 years ago

5 0

Answer:

5 (x - 1) (4 x + 7)

Step-by-step explanation:

Factor the following:

20 x^2 + 15 x - 35

Factor 5 out of 20 x^2 + 15 x - 35:

5 (4 x^2 + 3 x - 7)

Factor the quadratic 4 x^2 + 3 x - 7. The coefficient of x^2 is 4 and the constant term is -7. The product of 4 and -7 is -28. The factors of -28 which sum to 3 are -4 and 7. So 4 x^2 + 3 x - 7 = 4 x^2 + 7 x - 4 x - 7 = 7 (x - 1) + 4 x (x - 1):

5 7 (x - 1) + 4 x (x - 1)

Factor x - 1 from 7 (x - 1) + 4 x (x - 1):

Answer: 5 (x - 1) (4 x + 7)

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FinnZ [79.3K]

Answer:

110 students walk or bike to school.

Step-by-step explanation:

48% of the 208 students walk, whereas 5% bike. Combining 48% and 5%, we get 53%.

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2 years ago

Solve forx.<br> 0 = x2 + 8x + 15<br> Enter your answers in the boxes.<br> The solutions are | and

Allushta [10]

The anwser is 15 I think

6 0

1 year ago

Read 2 more answers

What is the value of a? Blank units

Damm [24]

Answer:

289

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a = 8 b= 15 c = s

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2 years ago

Solve the following equation for X. Round your answer to four decimal places.

strojnjashka [21]

ANSWER

x = 1.2226

EXPLANATION

To solve this equation we have to apply the property of the logarithm of the base,

$\log _bb=1$

Thus, we can apply the natural logarithm - whose base is e, to both sides of the equation,

$\ln e^{4x}=\ln 133$

Now we apply the property of the logarithm of a power,

$\log a^b=b\log a$

In our equation,

$\begin{gathered} 4x\ln e^{}=\ln 133 \\ 4x^{}=\ln 133 \end{gathered}$

Then divide both sides by 4 and solve,

$\begin{gathered} \frac{4x^{}}{4}=\frac{\ln 133}{4} \\ x\approx1.2226 \end{gathered}$

The solution to this equation is x = 1.2226, rounded to four decimal places.

4 0

1 year ago

Using the graph, determine the coordinates of the vertex of the parabola.

nadezda [96]

Answer:

The coordinates of the vertex is (6,4).

Step-by-step explanation:

The vertex is the highest or lowest point on the parabola (curved line).

6 0

2 years ago