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Readme [11.4K]
3 years ago
5

Can someone help me and tell me what I got wrong

Mathematics
2 answers:
aliina [53]3 years ago
6 0

I'm not sure what you did at first, but here's how I solved it:

7(3+3)-9

7(6)-9

42-9

=33

i can't see your original answer, so i can't tell you specifically what went wrong, but i hope that helps!

ira [324]3 years ago
4 0

Answer: 33

Step-by-step explanation:

f(a)=7(a+3)-9

f(3)=7(3+3)-9

f(3)=7(6)-9

f(3)=42-9

f(3)=33

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SAT question <br> Can someone explain to me what this is asking
frozen [14]

Answer:

B

Step-by-step explanation:

The vertex is the turning point of the parabola.

First find the zeros by equating y to zero, that is

(x - 6)(x + 12) = 0

Equate each factor to zero and solve for x

x - 6 = 0 ⇒ x = 6

x + 12 = 0 ⇒ x = - 12

The vertex lies on the axis of symmetry which is situated at the midpoint of the zeros, thus

x_{vertex} = \frac{6-12}{2} = \frac{-6}{2} = - 3

Thus the x- coordinate of the vertex is x = - 3 → B

6 0
4 years ago
Please help quick<br> due in 20 mins
IrinaK [193]

Answer:

5.7 I pretty sure

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Use Mental Math to find the value of 15/124 x 230 / 30 ÷ 230 / 124
aksik [14]
 it wold be this 0.000032518
6 0
3 years ago
Read 2 more answers
En una pension de perros hay tres perros que pesan 20kg 10kg y otro 5kg si se contempla que 70kg de alimento dura 2 meses¿que ca
aleksley [76]

Answer:

40kg

20kg

10kg

Step-by-step explanation:

The computation of the food each one is getting as follows

Total weights of all dogs

= 20 kg + 10 kg + 5kg

= 35 kg

And, the food is 70 kg

So, each one is getting

for 20 kg

= 70 ÷ 35 × 20

= 40 kg

For 10 kg

= 70 ÷ 35 × 10

= 20 kg

And, for 5 kg

= 70 ÷ 35 × 5

= 10 kg

7 0
3 years ago
How do you do these two questions?
nignag [31]

Step-by-step explanation:

(a) ∫₋ₒₒ°° f(x) dx

We can split this into three integrals:

= ∫₋ₒₒ⁻¹ f(x) dx + ∫₋₁¹ f(x) dx + ∫₁°° f(x) dx

Since the function is even (symmetrical about the y-axis), we can further simplify this as:

= ∫₋₁¹ f(x) dx + 2 ∫₁°° f(x) dx

The first integral is finite, so it converges.

For the second integral, we can use comparison test.

g(x) = e^(-½ x) is greater than f(x) = e^(-½ x²) for all x greater than 1.

We can show that g(x) converges:

∫₁°° e^(-½ x) dx = -2 e^(-½ x) |₁°° = -2 e^(-∞) − -2 e^(-½) = 0 + 2e^(-½).

Therefore, the smaller function f(x) also converges.

(b) The width of the intervals is:

Δx = (3 − -3) / 6 = 1

Evaluating the function at the beginning and end of each interval:

f(-3) = e^(-9/2)

f(-2) = e^(-2)

f(-1) = e^(-1/2)

f(0) = 1

f(1) = e^(-1/2)

f(2) = e^(-2)

f(3) = e^(-9/2)

Apply Simpson's rule:

S = Δx/3 [f(-3) + 4f(-2) + 2f(-1) + 4f(0) + 2f(1) + 4f(2) + f(3)]

S ≈ 2.5103

5 0
3 years ago
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