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3 years ago

2. 7x = 5 + 5x solve

Mathematics

1 answer:

Tresset [83]3 years ago

4 0

Answer:

x = 2 $\frac{1}{2}$ or $\frac{5}{2}$

Step-by-step explanation:

Subtract 5x from both sides. This leaves us with 2x = 5. Now divide both sides by 2 to isolate x. If you want an improper fraction you can stop at x = $\frac{5}{2}$ . If you want a mixed number you will have x = 2 $\frac{1}{2}$ .

Hope this helped! Please give Brainliest.

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Find the missing side or angle. Round to the nearest tenth. C=53° B= 80° a=2 b=[ ? ]

Degger [83]

9514 1404 393

Answer:

b ≈ 2.7

Step-by-step explanation:

The missing angle A is ...

A = 180° -53° -80° = 47°

The law of sines tells you ...

b/sin(B) = a/sin(A)

b = sin(80°)×2/sin(47°) ≈ 2.6931

Side b is about 2.7 units.

5 0

3 years ago

Read 2 more answers

Which equation is true when the value of x is -12

Sophie [7]

Answer:

Which equation is true when the value of x is -12 HERE YOU GO!!

Step-by-step explanation:

tricky ... let's see ...

I notice that if we subtract xy from both sides we get

7x + xy - xy = xy - xy + 21

then

7x = 21

and

x = 21/7 = 3

so there is only one value of x that satisfies the equation

x = 3

Going back to the original equation we see that any value of y will satisfy the original equation

we can see this by rearranging things:

7x + xy - xy = 21

here, I have performed the subtraction of xy on the right side as above, but have left the left side undone

(so we don't lose the presence of y)

Note that the above can also be written as

7x + (x - x)y = 21

7x + 0y =21

now, since anything times zero equals zero,

y may be any number.

Let's summarize:

1) x = 3

2) y = anything

looking back at the original question;

1) the equation is true for all ordered pairs

FALSE (only one x works, not all x)

2) there are no x and y pairs for which the equation is true.

FALSE (x=3, y=anything) makes it true, i.e. (3,1)

3) For each value of x, there is one and only one value of y that makes the equation true,

FALSE for each value of x, for the one value of x, x=3, y can be any number, which is an infinite number, not one

4) for each value of y, there is one and only on value of x that makes the equation true.

TRUE!! for all the infinite values of y you may pick, there is one and only one x you may pick, x = 3

ANSWER: STATEMENT 4 is CORRECT (TRUE)

3 0

2 years ago

The sum of three numbers is 8. The third is 9

Ierofanga [76]

Answer:

→ First value is 1

→ Second value is 2

→ Third value is 5

Step-by-step explanation:

• let numbers be x, y and z

${ \tt{z = 8y - 9 - - - (eqn \: 1)}} \\ \\ { \tt{10x = 8y - 7 - - - (eqn \: 2)}} \\ \\ { \tt{x + y + z = 8 - - - (eqn \: 3)}}$

• from eqn 2, make x the subject:

${ \tt{x = \frac{8y - 7}{10} }} \\$

• substitute all variables in eqn 3:

${ \tt{ \frac{8y - 7}{10} + y + 8y - 9 = 8}} \\ \\ { \tt{8y - 7 + 10y + 80y - 90 = 80}} \\ \\ { \tt{98y = 177}} \\ \\ { \boxed{ \tt{ \: y = 1.8}}}$

• find z

${ \tt{z = 8y - 9}} \\ \\ { \tt{z = 8(1.8) - 9}} \\ \\ { \tt{z = 14.4 - 9}} \\ \\ { \boxed{ \tt{ \: z = 5.4 \: }}}$

• find x:

${ \tt{x = \frac{8(1.8) - 7}{10} }} \\ \\ { \tt{x = \frac{7.4}{10} }} \\ \\ { \boxed{ \tt{ \: x = 0.74 \: }}}$

Rounding to nearest value:

${ \boxed{ \rm{x = 1}}} \\ { \boxed{ \rm{y =2 }}} \\ { \boxed{ \rm{z = 5}}}$

7 0

2 years ago

Find the area and perimeter of a rectangle if the length is 25 and the width is 8.

g100num [7]

Answer:

area is 200 perimeter is 66

Step-by-step explanation:

3 0

2 years ago

Suppose a consumer product researcher wanted to find out whether a highlighter lasted less than the manufacturer's claim that th

11111nata11111 [884]

Answer:

The null hypothesis is $H_0: \mu = 14$

The alternate hypothesis is $H_a: \mu < 14$

The test statistic is t = -1.95.

The p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the highlighters wrote for less than 14 continuous hours.

Step-by-step explanation:

Suppose a consumer product researcher wanted to find out whether a highlighter lasted less than the manufacturer's claim that their highlighters could write continuously for 14 hours.

At the null hypothesis, we test if the mean is 14 hours, that is:

$H_0: \mu = 14$

At the alternate hypothesis, we test if the mean is less than 14 hours, that is:

$H_a: \mu < 14$

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.

14 is tested at the null hypothesis:

This means that $\mu = 14$

X = 13.6 hours, s = 1.3 hours. Sample of 40:

In addition to the values of X and s given, we have that $n = 40$

Test statistic:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{13.6 - 14}{\frac{1.3}{\sqrt{40}}}$

$t = -1.95$

The test statistic is t = -1.95.

P-value:

The p-value of the test is the probability of finding a sample mean lower than 13.6, which is a left tailed test, with t = -1.95 and 40 - 1 = 39 degrees of freedom.

Using a calculator, the p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the highlighters wrote for less than 14 continuous hours.

5 0

2 years ago