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2 years ago

X + 11 = 3 so the equation

Mathematics

1 answer:

Mnenie [13.5K]2 years ago

8 0

X+11=3
-11 -11
x=-8

you subtract 11 from both sides. then you got the answer.

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Help please, I attached the question. Is it a!?

AVprozaik [17]

Answer:

Step-by-step explanation:

Recall that for a quadratic equation of the form:

$0=ax^2+bx+c$

The number of solutions it has can be determined using its discriminant:

$\Delta = b^2-4ac$

Where:

If the discriminant is positive, we have two real solutions.
If the discriminant is negative, we have no real solutions.
And if the discriminant is zero, we have exactly one solution.

We have the equation:

$2x^2+5x-k=0$

Thus, a = 2, b = 5, and c = -k.

In order for the equation to have exactly one distinct solution, the discriminant must equal zero. Hence:

$b^2-4ac=0$

Substitute:

$(5)^2-4(2)(-k)=0$

Solve for k. Simplify:

$25+8k=0$

Solve:

$\displaystyle k = -\frac{25}{8}$

Thus, our answer is indeed A.

3 0

2 years ago

Read 2 more answers

Solve for the value of x. ABC 20 28 9x-8

Marizza181 [45]

28+20+9x-8=180

48+9x-8=180

9x=180-40

9x=140

×=15.56

6 0

2 years ago

Please help me with this problem

deff fn [24]

Answer:

C. -3(√y)(∛x²)

Step-by-step explanation:

Recall the rule for exponents with fractions. The numerator represents the power the variable is raised by, and the denominator represents the which root will be taken.

y^1/2 equals √y

and x^2/3 equals ∛x²

So the answer is -3(√y)(∛x²)

Look at the attached image for more clarification

7 0

2 years ago

Which polynomial is in standard form?

skelet666 [1.2K]

Answer:

Step-by-step explanation:

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3 years ago

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Hello can someone help me simplify 4) and 5)? Thank you and please include steps :)

Leni [432]

Alrighty

remember some rules
$(ab)^c=(a^c)(b^c)$
and
$x^{-m}=\frac{1}{x^m}$
and
$x^0=1$ for all real values of x
and
$(a^b)^c=a^{bc}$
and
$(\frac{a}{b})^c=\frac{a^c}{b^c}$
and
(a/b)/(c/d)=(ad)/(bc)
and
$(a^b)(a^c)=a^{b+c}$
and
$\frac{a^m}{a^n}=a^{m-n}$
and don't forget pemdas
example: $-x^m=-1(x^m)$ but $(-x)^m=(-1)^m(x^m)$

so

4.
$(\frac{c^{-2}}{2})^{-2}=$

$\frac{(c^{-2})^{-2}}{2^{-2}}=$

$\frac{c^4}{\frac{1}{2^2}}=$

$\frac{c^4}{\frac{1}{4}}=$

$4c^4$

5.

$\frac{(-a)^4bc^5}{-a^2b^{-3}c^0}=$

$\frac{(-1)^4(a)^4bc^5}{-1(a^2)(\frac{1}{b^3}(1)}=$

$\frac{(1)a^4bc^5}{\frac{(-1)(a^2)}{b^3}}=$

$\frac{(a^4bc^5)b^3}{(-1)(a^2)}=$

$\frac{-a^4b^4c^5}{a^2}=$

$-a^2b^4c^5$

3 0

3 years ago