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3 years ago

Pleaseee, I really need help with this❤️

Mathematics

1 answer:

Bess [88]3 years ago

7 0

Answer:

(4x + 5)(4x - 3)

Step-by-step explanation:

Given

16x² + 8x - 15

Consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term.

product = 16 × - 15 = - 240 and sum = + 8

The factors are + 20 and - 12

Use these factors to split the x- term

16x² + 20x - 12x - 15 ( factor the first/second and third/fourth terms )

= 4x(4x + 5) - 3(4x + 5) ← factor out (4x + 5) from each term

= (4x + 5)(4x - 3) ← in factored form

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Tom went bowling with $27 to spend.He rented shoes for $5.75 and paid $4.25 for each game.

RideAnS [48]

Answer:

Step-by-step explanation:

Subtract the spent money to rent cost:

27-5.75=21.25

Divide the answer to the cost of each game:

21.25/4.25=5

he played 5 times.

6 0

3 years ago

a shipping company charged $15 to ship a package and 5% to ensure the package what was total cost to ship this

postnew [5]

Answer:

15.75

Step-by-step explanation:

So we know 5% is 1/20 out of 100 (20x5=100) so then we need to set up the equation to find how much 5% of 15 is. Which is .75 cents. So we then add the 75 cents to the already $15 shipping fee to get an answer of 15 dollars and seventy five cents.

6 0

3 years ago

Read 2 more answers

The value of 4 in 46,395 is how many times the value of 4 in 14,906?

serg [7]

The 4 in 46,395 is 10 times the 4 in 14,906.

5 0

3 years ago

Read 2 more answers

The length and width of a rectangle are measured as 55 cm and 49 cm, respectively, with an error in measurement of at most 0.1 c

valentina_108 [34]

Answer:

The maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

Step-by-step explanation:

The area of a rectangle with length $L$ and width $W$ is $A= L\cdot W$ so the differential of <em>A</em> is

$dA=\frac{\partial A}{\partial L} \Delta L+\frac{\partial A}{\partial W} \Delta W$

$\frac{\partial A}{\partial L} = W\\\frac{\partial A}{\partial W}=L$ so

$dA=W\Delta L+L \Delta W$

We know that each error is at most 0.1 cm, we have $|\Delta L|\leq 0.1$ , $|\Delta W|\leq 0.1$ . To find the maximum error in the calculated area of the rectangle we take $\Delta L = 0.1$ , $\Delta W = 0.1$ and $L=55$ , $W=49$ . This gives

$dA=49\cdot 0.1+55 \cdot 0.1$

$dA=10.4$

Thus the maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

4 0

3 years ago

Help me, please!!!!! CORRECT ANSWER GETS BRAINLIEST!!!!!!

zavuch27 [327]

Hello,

Here is your answer:

The proper answer to your question is..... 200,000=4x-20

When your doing questions like these make sure you pay attention to the question.

Your answer is 200,000=4x-20

If you need anymore help feel free to ask me!

Hope this helps!

7 0

3 years ago