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Anon25 [30]
4 years ago
11

2(x + 2) + 3x = 2(x + 1) + 1

Mathematics
1 answer:
monitta4 years ago
4 0

Answer:

X= -1/3

Step-by-step explanation:

In the picture it shows the work that i did

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assuming all parabolas are of the form y = ax^2 + bx + c drag and drop the graphs to match the appropriate a-value
Maslowich

Answer:

A parabola has the form:

y = a*x^2 + b*x + c

Where a is the leading coefficient.

If a is positive, the parabola opens up.

If a is negative, the parabola opens down.

a is the factor that multiplies the part that grows the fastest in the equation, thus, if a is a larger value (in absolute value) then the parabola will grow faster (then the parabola will be narrow)

if a is smaller (again, in absolute value) the parabola will grow slower, then the parabola will be wider.

With this, we can conclude that:

a = -4

is the largest value of a in absolute value.

Then this corresponds to the thinner parabola (the one at the left)

a = -1

Is the middle value of a, then this corresponds to the graph of the middle

a = -0.25

Is the smallest absolute value of a, then this one corresponds to the widest graph (the first one at the left)

5 0
3 years ago
Please help
lisov135 [29]

Answer: 12

Step-by-step explanation:

law of cosines states that f^2 = t^2 + p^2 - 2tp cos(F)

so f^2 = 5^2 + 7^2 - 2*5*7*cos(175)

f^2 = 74 - (-69.7336)

f^2 = 143.7336

f = 11.9889

which rounds to 12

7 0
3 years ago
Read 2 more answers
Divide 33 photos into two groups so the ratio is 4 to 7
Alik [6]
Group 1: 12
Group 2: 21
7 0
3 years ago
V36 What smallest number should be multiplied to 288 in order to make it a perfect cube?​
olga55 [171]

Answer:

6

Step-by-step explanation: We observe that if 288 is multiplied by (2 x 3), then its prime factors will exist in triples. Thus, the required smallest number by which 288 be multiplied to make it a perfect cube i: (2 x 3)=6. 1,331 is a perfect cube.

6 0
3 years ago
7 1/2 divided by 1 1/2 = 5 How to check the answer
Gnom [1K]

Answer:

We verified that the equation 7\frac{1}{2}\div1\frac{1}{2}=5 has both sides equal.

Step-by-step explanation:

We need to solve and verify: 7\frac{1}{2}\div1\frac{1}{2}=5

Solving:

7\frac{1}{2}\div1\frac{1}{2}=5

Converting into improper fractions

\frac{15}{2}\div\frac{3}{2}=5

Converting division sign into multiplication and reversing 3/2 i.e

\frac{15}{2}\times\frac{2}{3}=5

Now simplifying

\frac{15}{3}=5 \\5=5

So, We verified that the equation 7\frac{1}{2}\div1\frac{1}{2}=5 has both sides equal.

7 0
3 years ago
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