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Mrrafil [7]
3 years ago
10

What is the standard form for this equation (x+2) (x+15)

Mathematics
1 answer:
MissTica3 years ago
4 0

Answer:

x^2 + 17x + 30

Step-by-step explanation:

Just multiply the components from first bracket to components of second bracket, then add like terms and rearrange to write numbers from highest exponents to lowest.

You might be interested in
If an object is propelled upward from a height of 72 feet at an initial velocity of 90 feet per second, then its height h after
kifflom [539]

Answer:

6.34 seconds.

Step-by-step explanation:

The object will hit the ground when h = 0.

-16t^2 + 90t + 72 = 0

8t^2 - 45t - 36 = 0

We can then use the quadratic formula to solve.

[please ignore the A-hat; that is a bug]

\frac{45±\sqrt{45^2 - 4 * 8 * -36} }{2 * 8}

= \frac{45±\sqrt{2025 + 1152} }{16}

= \frac{45±\sqrt{3177} }{16}

= \frac{45±56.36488268}{16}

(45 - 56.36488268) / 16 = -0.7103051678

(45 + 56.36488268) / 16 = 6.335305168

Since the time cannot be negative, the object will hit the ground after about 6.34 seconds.

Hope this helps!

4 0
3 years ago
Find the 100th term of an=17-4(n-1)
egoroff_w [7]

Answer:

Find the 100th term of an=17-4(n-1)

Step-by-step explanation:

Arithmetic sequence : -17, -14, -11, -8, ...

Here the first term , a 1 = -17

Common difference :    d = -14 - (-17) =3

4 0
2 years ago
Read 2 more answers
Solve the equation for x 25=1/x​
Mkey [24]

Answer:

x = 1/25

Step-by-step explanation:

Solve for x:

25 = 1/x

25 = 1/x is equivalent to 1/x = 25:

1/x = 25

Take the reciprocal of both sides:

Answer:  x = 1/25

5 0
3 years ago
A contractor cut 28 feet of wire into sections that were 4 2/3 feet long. How many pieces of wire did the contractor have?
Alisiya [41]
6

28 divided by 4 2/3=6
:)
3 0
3 years ago
Solve 2x + 4 &gt; 16<br> A. X &lt; 6<br> B. X &gt; 6<br> C. X &lt; 10<br> D. X &gt; 10
Korvikt [17]

Answer:

B

2(6)+4 > 16             x > 6                 2(7) + 4 > 16

12+4=16                                            14+4>16       18>16

8 0
4 years ago
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