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2 years ago

10x = -71 - 11y3x = -7 + 11y

Mathematics

2 answers:

lyudmila [28]2 years ago

6 0

(-6,-1) is the answer to your question.

WINSTONCH [101]2 years ago

5 0

If we add theses 2 equations we eliminate the y terms ( -11y + 11y = 0):-

so adding we get

13x = -78

x = -6

Substitute x = -6 in the first equation:-

10(-6) = -71-11y

11y = -71 + 60 = -11

y = 1

So the solution to the system of equations is x = -6,y = 1

You might be interested in

Lf x+1/x=12 then find the value of x3+1/x3

Tresset [83]

Answer:

Explanation:

First we find what x is:
x + 1/x = 12
x + 1 = 12x
1 = 12x - x
1 = 11x
1/11 = x
Or x = 1/11

Plug x value in x^3 + 1/x^3

(1/11)^3 + 1/(1/11)^3
= (1^3/11^3)+ 1/(1^3/11^3)
= (1/1331 + 1)/1/1331
= (1/1331 + 1331/1331)/1/1331
= 1332/1331 x 1331/1
= 1332/1
= 1332

Therefore, x^3 + 1/x^3 = 1332

4 0

2 years ago

Brandi solved 16 / 5 by using a related multiplication expression. What multiplication expression did she is?

klemol [59]

The given question is wrong.

Question:

Brandi solved $\frac{1}{6}\div 5$ by using a related multiplication expression. What multiplication expression did she is?

Answer:

The related multiplication expression is $$\frac{1}{6}\times \frac{1}{5}$ .

Solution:

Given expression is $\frac{1}{6}\div 5$ .

We can write 5 as a fraction with denominator 1.

$$ \frac{1}{6}\div5=\frac{1}{6}\div \frac{5}{1}$

Using the fraction rule $$\frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}$ ,

$$\frac{1}{6}\div \frac{5}{1}=\frac{1}{6}\times \frac{1}{5}$

Now multiply the fractions using the rule $$\frac{a}{b}\times \frac{d}{c}=\frac{a\times d}{b\times c}$ ,

$$\frac{1}{6}\times \frac{1}{5}=\frac{1\times1}{6\times5}$

Let us multiply the numbers in the numerator and denominator separately, we get

$$\frac{1\times1}{6\times5}=\frac{1}{30}$

Hence the related multiplication expression is $$\frac{1}{6}\times \frac{1}{5}$ .

8 0

2 years ago

What is the length of line segment PQ? If you can please explain, thanks.

nydimaria [60]

Given:

Tangent segment MN = 6

External segment NQ = 4

Secant segment NP =x + 4

To find:

The length of line segment PQ.

Solution:

Property of tangent and secant segment:

If a secant and a tangent intersect outside a circle, then the product of the secant segment and external segment is equal to the product of the tangent segment.

$\Rightarrow NQ\times NP = MN^2$

$\Rightarrow 4\times (x+4) = 6^2$

$\Rightarrow 4x+16 = 36$

Subtract 16 from both sides.

$\Rightarrow 4x+16-16 = 36-16$

$\Rightarrow 4x =20$

Divide by 4 on both sides.

$$\Rightarrow\frac{4x}{4}=\frac{20}{4}$

$\Rightarrow x = 5$

The length of line segment PQ is 5 units.

5 0

3 years ago

Read 2 more answers

A large university accepts 60% of the students who apply. Of the students the university accepts, 40% actually enroll. If 30,

kykrilka [37]

Answer:

7,200

Step-by-step explanation:

Let's start by calculating how many students the university accepts. The problem says they accept 60% of the students. 60% as a decimal is 0.6. To calculate 60% of something, you multiply .60 by whatever quantity. In this case, it would be the students.

Number of accepted students = $.6 * 30000$ = 18,000

The university accepts 18,000 students.

Now, the next problem is that of those accepted students, we need to find out how many actually enroll. The problem says they accept 40% of the accepted students. Applying the same principle,

Number of enrolled students = $.4 * 18000$ = 7,200