Ask question

Ask question

All categories

2 years ago

11

PLEASE HELP!!!

2 answers:

Jet001 [13]2 years ago

6 0

Answer:

see file below.

Step-by-step explanation:

Akimi4 [234]2 years ago

4 0

Answer:

he has a lala ramjan I am not sure how are the true meaning of life and the heart ☺️ to

You might be interested in

The mean monthly gasoline bill for one household is greater than $140. If a hypothesis test is performed, how should you inter

adell [148]

Answer: C. There is sufficient evidence to support the claim . $\mu>140$

Step-by-step explanation:

Given (claim) :The mean monthly gasoline bill for one household is greater than $140.

Let $\mu$ be the population mean.

Then, the set of hypothesis for this claim will be:-

$H_0:\mu\leq140\\\\ H_a:\mu>140$

If a hypothesis test is performed , then to interpret a decision that fails to reject the null hypothesis we says that there is sufficient evidence to support the claim $\mu>140$

6 0

3 years ago

Can anyone share the process please?

gtnhenbr [62]

Answer:

3x³ + x² + x + 1

= 3x³ - 3x² + 4x² - 4x + 5x - 5 + 6

= 3x²(x - 1) + 4x(x - 1) + 5(x - 1) + 6

= (x - 1)(3x² + 4x + 5) + 6

but 3x³ + x² + x + 1 = (x - 1).Q(x) + R

=> Q(x) = 3x² + 4x + 5

R = 6

4 0

2 years ago

Find the product (-3) • (1.2)<br><br> A. 4.5 <br><br> B. -3.6<br><br> C. -4.6<br><br> D. 5.3

NikAS [45]

The answer is b -3.6 hope this helps and god bless my friend

4 0

3 years ago

Read 2 more answers

Determine whether each expression is equivalent to 49^2t – 0.5.

vampirchik [111]

Answer:

None of the expression are equivalent to $49^{(2t - 0.5)}$

Step-by-step explanation:

Given

$49^{(2t - 0.5)}$

Required

Find its equivalents

We start by expanding the given expression

$49^{(2t - 0.5)}$

Expand 49

$(7^2)^{(2t - 0.5)}$

$7^2^{(2t - 0.5)}$

Using laws of indices: $(a^m)^n = a^{mn}$

$7^{(2*2t - 2*0.5)}$

$7^{(4t - 1)}$

This implies that; each of the following options A,B and C must be equivalent to $49^{(2t - 0.5)}$ or alternatively, $7^{(4t - 1)}$

A. $\frac{7^{2t}}{49^{0.5}}$

Using law of indices which states;

$a^{mn} = (a^m)^n$

Applying this law to the numerator; we have

$\frac{(7^{2})^{t}}{49^{0.5}}$

Expand expression in bracket

$\frac{(7 * 7)^{t}}{49^{0.5}}$

$\frac{49^{t}}{49^{0.5}}$

Also; Using law of indices which states;

$\frac{a^{m}}{a^n} = a^{m-n}$

$\frac{49^{t}}{49^{0.5}}$ becomes

$49^{t-0.5}}$

This is not equivalent to $49^{(2t - 0.5)}$

B. $\frac{49^{2t}}{7^{0.5}}$

Expand numerator

$\frac{(7*7)^{2t}}{7^{0.5}}$

$\frac{(7^2)^{2t}}{7^{0.5}}$

Using law of indices which states;

$(a^m)^n = a^{mn}$

Applying this law to the numerator; we have

$\frac{7^{2*2t}}{7^{0.5}}$

$\frac{7^{4t}}{7^{0.5}}$

Also; Using law of indices which states;

$\frac{a^{m}}{a^n} = a^{m-n}$

$\frac{7^{4t}}{7^{0.5}}$ = $7^{4t - 0.5}$

This is also not equivalent to $49^{(2t - 0.5)}$

C. $7^{2t}\ *\ 49^{0.5}$

$7^{2t}\ *\ (7^2)^{0.5}$

$7^{2t}\ *\ 7^{2*0.5}$

$7^{2t}\ *\ 7^{1}$

Using law of indices which states;

$a^m*a^n = a^{m+n}$

$7^{2t+ 1}$

This is also not equivalent to $49^{(2t - 0.5)}$

6 0

3 years ago

Solve this equation. Enter your answer in the box.<br> 13(y+7)=3(y−1).

Fofino [41]

★To Solve :

$\rm\blue{13(y+7) = 3(y-1)}$

⠀⠀⠀⠀⠀

★SOLUTION :

$\rm\red\leadsto{13(y+7) = 3(y-1)}$

$\rm\red\leadsto{13\times{y}+13\times7= 3\times{y}-3\times1}$

$\rm\red\leadsto{13y+91=3y-3}$

$\rm\red\leadsto{13y-3y=-3-91}$

$\rm\red\leadsto{10y=-94}$

$\rm\red\leadsto{} y =\dfrac{ { \cancel{-94}}^{ \:-45 } }{ { \cancel{10}}^{ \: 5} }$

$\rm\red\leadsto{y=}\dfrac{-47}{5}$

____________________

5 0

2 years ago

Read 2 more answers