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3 years ago

Describe how factoring a quadratic expression ax2 + bx + c, where a ≠ 1, is different from factoring x2 + bx + c.

Mathematics

1 answer:

garik1379 [7]3 years ago

6 0

Answer:

sjjrnf b

Step-by-step explanation:

bbjjdjmend. fnrm xx

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What would the new point be If you rotated the point (3,-5) 270 clockwise.

timurjin [86]

Answer:

see below

Step-by-step explanation:

When you rotate a point 270 degrees clockwise (x, -y) becomes (-y, -x) so the answer would be (-5, -3).

5 0

3 years ago

A wallet is originally $11.00. With a 40% discount, what is the sales price?

Leviafan [203]

Answer:

$6.60

Step-by-step explanation:

11 x .40 = 4.4

11 - 4.4 = 6.6

6 0

3 years ago

Read 2 more answers

Consider this expression.

Oliga [24]

A
D
F
That is the correct answer

4 0

3 years ago

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Christov and Mateus gave out candy to children on Halloween. They each have out candy at a constant rate, and they both gave awa

padilas [110]

Answer: 1) Both gave same number of candies to each child.

2) Christov gave candy to a greater number of children.

Step-by-step explanation:

Since, Christov initially had 300 pieces of candy, and after he was visited by 17 children, he had 249 pieces left.

Let he gave x candy to each student when he visited 17 children.

Then, 17 x + 249 = 300

⇒ 17 x = 51

⇒ x = 3

Since, he distributes candies in a constant speed.

Therefore, his speed of distribution = 3 candies per child

Now,The function that shows the remaining number of candies Mateus has after distributing candies to n children,

C(n)=270 - 3 n

Initially, n = 0

C(0) = 270

⇒ Mateus has initial number of candies = 270

When he gave candies to one child then remaining candies = 270 - 3 × 1 = 267

Thus, the candies, get by a child from Mateus = 270 - 267 = 3

Since, he distributes candies in a constant speed.

⇒ His speed of distribution = 3 candies per child

1) Therefore, Both Christov and Mateus have same speed of distribution.

2) Since, both have same seed.

⇒ The one who has greater number of candies will be distribute more.

⇒ Christov will give more candies.

8 0

4 years ago

What is the third quartile, Q3, of the following distribution?

GREYUIT [131]

Answer:

The third quartile is:

$Q_3=29$

Step-by-step explanation:

First organize the data from lowest to highest

4, 5, 10, 12, 14, 16, 18, 20, 21, 21, 22, 22, 24, 26, 29, 29, 33, 34, 43, 44

Notice that we have a quantity of n = 20 data

Use the following formula to calculate the third quartile $Q_3$

For a set of n data organized in the form:

$x_1, x_2, x_3, ..., x_n$

The third quartile is $Q_3$ :

$Q_3=x_{\frac{3}{4}(n+1)}$

With n=20

$Q_3=x_{\frac{3}{4}(20+1)}$

$Q_3=x_{15.75}$

The third quartile is between $x_{15}=29$ and $x_{16}=29$

Then

$Q_3 =x_{15} + 0.75*(x_{16}- x_{15})$

$Q_3 =29 + 0.75*(29- 29)\\\\Q_3 =29$

5 0

4 years ago

Read 2 more answers