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3 years ago

If x=-3 and y=-2 Please help me

Mathematics

2 answers:

guapka [62]3 years ago

8 0

Answer:

-33

Step-by-step explanation:

Dimas [21]3 years ago

6 0

Answer:

-15

Step-by-step explanation:

(-3)^2 - 3(-3)(-2) + 2(-3)

You might be interested in

What is the discriminante of 9x^2+10x

Leni [432]

The discriminant of y=ax^2+bx+c is b^2-4ac

given
9x^2+10x+0
a=9
b=10
c=0

discriminant=10^2-4(9)(0)=100-0=100

the discriminant is 100

6 0

3 years ago

Bob's Bagels sells 7 bagels for $8.00, and Betty's Bagels sells 9 bagels

puteri [66]

Answer:

3 of 4

The better deal will have the higher ratio of dollars per bagel.

To find the better deal, you must compare the fractions 8/7 and 6/9.

Comparng the fractions reveals that Betty has the better deal

Step-by-step explanation:

bobbs bagels: 8/7 equals 1.14, so each bagel would cost about $1.14

bettys bagels: 6/9 equals 0.66, so each bagel would cost about $0.66

$0.66 is better than $1.14 per bagel. therefore, bettys bagels are cheaper

hope this helps :)

7 0

2 years ago

Write each number as a product of a decimal number between 1 and 10 and a power of 10. 234,000

MrRa [10]

Answer:

2.34 * 10^5.

Step-by-step explanation:

234000 = 2.34 * 10^5

- there are 5 digits after the first number so the exponent is 5.

4 0

3 years ago

Carina says that to divide by 12, she will need to write 12 as the fraction and then divide. Is she correct? Explain your reason

Alika [10]

Answer:

I do not know

Step-by-step explanation:

I think so

6 0

2 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Andre45 [30]

For this case, we must indicate which of the given functions is not defined for $x = 0$

By definition, we know that:

$f (x) = \sqrt {x}$ has a domain from 0 to infinity.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .

While $y = \sqrt {x + 2}$ if it is defined for $x = 0.$

$f(x)=\sqrt[3]{x}$ , your domain is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root.

So, we have:

$y = \sqrt [3] {x-2}$ with x = 0: $y = \sqrt [3] {- 2}$ is defined.

$y = \sqrt [3] {x + 2}$ with x = 0: $y = \sqrt [3] {2}$ in the same way is defined.

Answer:

$y = \sqrt {x-2}$

Option b

6 0

3 years ago