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3 years ago

Solve in polynomials (3/4)^-2 simplify the expression

Mathematics

1 answer:

valentina_108 [34]3 years ago

6 0

$(\frac{3}{4})^{-2} = \frac{3^{-2}}{4^{-2}} = \frac{4^2}{3^2} = \frac{16}{9} = 1.777...$

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What is the factored form of 11x-x^2=0

Aleks [24]

Answer: $x(11-x)=0$

Step-by-step explanation:

Given the quadratic equation $11x-x^2=0$ , you need to factor it.

In order to find the form asked of the given equation, you need to factor out the common factor of the terms.

You can observe that the common factor of the terms of the equation is: $x$

Now, knowing this, you must factor out $x$ . Then you get the following form:

$x(11-x)=0$

Therefore, the factored fom of the equation $11x-x^2=0$ is:

$x(11-x)=0$

3 0

3 years ago

Solve the equation.

nadya68 [22]

Answer:

A. No solution

Step-by-step explanation:

The given equation is 4(2 x + 9) = 4(2 x +6)

We divide through by 4 to get:

2 x + 9 = 2 x +6

We now group similar terms to obtain:

$2x - 2x = 6 - 9$

$0 = - 3$

This final statement is false.

Therefore the given equation has no solution.

7 0

3 years ago

Help with this pleaseeee

ELEN [110]

Answer:

(- 3, 37) and (- $\frac{5}{2}$ , $\frac{63}{2}$ )

Step-by-step explanation:

Given the 2 equations

2x² - y + 19 = 0 → (1)

y + 11x = 4 → (2) ← subtract 11x from both sides

y = 4 - 11x → (3)

Substitute y = 4 - 11x into (1)

2x² - (4 - 11x) + 19 = 0

2x² - 4 + 11x + 19 = 0

2x² + 11x + 15 = 0 ← in standard form

(2x + 5)(x + 3) = 0 ← in factored form

Equate each factor to zero and solve for x

2x + 5 = 0 ⇒ 2x = - 5 ⇒ x = - $\frac{5}{2}$

x + 3 = 0 ⇒ x = - 3

Substitute these values into (3) for corresponding values of y

x = - $\frac{5}{2}$ : y = 4 + $\frac{55}{2}$ = $\frac{63}{2}$ ⇒ (- $\frac{5}{2}$ , $\frac{63}{2}$ )

x = - 3 : y = 4 + 33 = 37 ⇒ (- 3, 37 )

4 0

2 years ago

Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.

lorasvet [3.4K]

Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

$PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ$

Likewise:

$\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ$

Since they all measure 60°.

Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

$m\angle PAC=m\angle PAB+m\angle BAC$

Likewise:

$m\angle QAB=m\angle QAC+m\angle BAC$

Since ∠QAC ≅ ∠PAB:

$m\angle PAC=m\angle QAC+m\angle BAC$

And by substitution:

$m\angle PAC=m\angle QAB$

Thus:

$\angle PAC\cong \angle QAB$

Then by SAS Congruence:

$\Delta PAC\cong \Delta BAQ$

And by CPCTC:

$PC\cong BQ$

5 0

2 years ago

Read 2 more answers

8th grade math help please zoom in on the question for a better look (will give brainliest and thanks if it's right)

eduard

The two pairs are Pythagorean triples because if you plug the two legs of a right triangle into the Pythagorean theorem(A^2+B^2=C^2), then you will find the measurement for the third side(hypotenuse). i.e. 15^2+12^2=9^2(it's the Pythagorean triple 3,4,5 multiplied by 3). This works with any triple, as long as your using the legs of the triangle and as long as the triangle is a right triangle.

5 0

3 years ago