which expression is equavilent to 4/5x+(-2)+(-6)-1/5x​

Answer:

I don't understand please elaborate more your given question.

Step-by-step explanation:

Answer:

95m

Step-by-step explanation:

312 - 61 - 61 = 190

190/2 = 95

Answer:

a) 2 * 3 = 6.

b) 123 is odd but contains an even digit (2).

Step-by-step explanation:

Answer:

What is the common ratio of the sequence 3, 21, 147, …?  

7

Which formula can be used to find the nth term of the sequence 3, 21, 147, …?

c)

Use the given formula to find the indicated terms of the sequence.

a4 = 1029

a5 = 7203

Step-by-step explanation:

Answers for all three questions<3

Answer: 9 triangles

Step-by-step explanation:

Answer:

ok so.u grit da he on 40/ rock cause u got a andriodnh on gf

Answer:

x = -3

Step-by-step explanation:

Answer:

x=-3

Step-by-step explanation:

You can just do +x on both sides so: 0=3+x and then -3 on both sides: -3=x.

Answer: i think it is y=−56x−4

Step-by-step explanation:

The equation in the slope intercept form which passes through the points ( 0, -4 ) and ( 6 , 9 ) is y = (-5 / 6)x - 4.

The correct answer is Option C.

Given data:

To find the equation of a line in slope-intercept form (y = mx + b) that passes through the points (0, -4) and (6, -9), we need to determine the slope (m) and the y-intercept (b).

First, calculate the slope (m):

m = (change in y) / (change in x)

m = (-9 - (-4)) / (6 - 0)

m = (-9 + 4) / 6

m = -5 / 6

Now that we have the slope, we can use one of the given points (let's use (0, -4)) to solve for the y-intercept (b):

-4 = (-5 / 6) * 0 + b

-4 = b

So, the y-intercept (b) is -4.

Now, we can write the equation of the line in slope-intercept form:

y = (-5 / 6)x - 4

Hence, the equation of the line is y = (-5 / 6)x - 4.

To learn more about equation of line, refer:

https://brainly.com/question/14200719

#SPJ3

The complete question is attached below:

Which equation, in slope-intercept form, matches the equation shown?

a line that goes through the points (0, -4) and (6, -9)

A) y = ( 4/7 )x - 4

B) y = ( 5/6 )x + 1

C) y = ( -5/6 )x - 4

D) y = ( -4/7 )x + 1

Answer:

After 5 additional shots, the brothers will have a tied score.

Step-by-step explanation:

find equations

henry:

y-int is 15 because that is the starting number. the slope is 3 because that is how much his points are increasing for every successful shot. we can use this information to form this equation:

y = 3x + 15

brother:

y-int is 20 because that is the starting number. slope is 3 because that is how many points he gets for each successful shot. using this information, we can form the following equation:

y = 2x + 20

--

solve

y = 3x + 15

y = 2x + 20

3x + 15 = 2x + 20

x + 15 = 20

x = 5

--

graph

(just use desmos)

the answer is -2w (i have the write at least 20 words to dont worry about this)

Answer:

-2w

Step-by-step explanation:

Like terms have same variable.

34w - 4g + 4g - 36w = 34w - 36w - 4g + 4g

                                  = - 2w + 0 = -2w

y = 3x +4

3 is the slope. Rise 3 run 1 or up 3 right 1.

4 is the y-intercept

Answer:

z

Step-by-step explanation:

hqvqhqvw karrar ras wallah a part

Answer:

-1/6

Step-by-step explanation:

apply fraction rule to get 1/2(-3)

Multiply the numbers to get 1/-6

apply fraction rule to move negative to make

-1/6

1/2 divided by (-3) = -1/6 or -0.1666666667.

Answer:

The first three are functions

Step-by-step explanation:

4th one fails the Vertical line test.

Answer:

Option D)  [tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex]

Step-by-step explanation:

Given the equation, T = 15 + c × n, where:

T = represents the total monthly gym cost

c =  represents the additional charge for a group class, and

n = represents the number of group classes attended

In order to determine which equation can be used to find the number of group classes a customer attended, if there are given values for c and T, we must isolate the variable, n  algebraically.

The first step is to subtract 15 from both sides:

T = 15 + c × n

T - 15 = 15 - 15 + c × n

T - 15 = c × n

Next, divide both sides by c to isolate n :

[tex]\huge\mathsf{\frac{({T\:-\:15})}{c}\:=\:\frac{{c\:\times\:n}}{c}}[/tex]

[tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex]

Therefore, the correct answer is Option D) [tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex].

Step-by-step explanation:

je........porte

tu.......portes

nous.......portons

vous........portez

ils......portent

Step-by-step explanation:

[tex] = {( \frac{27}{8} )}^{ \frac{1}{3} } \times ( \frac{243}{32} )^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]

[tex] = { ({ (\frac{3}{2} )}^{3}) }^{ \frac{1}{3} } \times {( {( \frac{3}{2}) }^{5} )}^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]

[tex] = {( \frac{3}{2} )}^{3 \times \frac{1}{3} } \times {( \frac{3}{2} )}^{5 \times \frac{1}{5} } \times {( \frac{3}{2} )}^{2} [/tex]

[tex] = \frac{3}{2} \times \frac{3}{2} \times {( \frac{3}{2} )}^{2} [/tex]

[tex] = {( \frac{3}{2} )}^{1 + 1 + 2} [/tex]

[tex] = {( \frac{3}{2} )}^{4} \: or \: \frac{81}{16} [/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{27}{8} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{243}{32} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

We can write as :

27 = 3 × 3 × 3 = 3³

8 = 2 × 2 × 2 = 2³

243 = 3 × 3 × 3 × 3 × 3 = 3⁵

32 = 2 × 2 × 2 ×2 × 2 = 2⁵

[tex]\sf{\longmapsto{\bigg( \dfrac{3 \times 3 \times 3}{2 \times 2 \times 2} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{{(3)}^{3}}{{(2)}^{3}} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{({3}^{5})}{{(2)}^{5}} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

Now, we can write as :

(3³/2³) = (3/2)³

(3⁵/2⁵) = (3/2)⁵

[tex]\sf{\longmapsto{\left\{\bigg(\frac{3}{2} \bigg)^{3} \right\}^{\frac{1}{3}} \times \Bigg[\left\{\bigg(\frac{3}{2} \bigg)^{5} \right\}^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

Now using law of exponent :

[tex]{\sf{({a}^{m})^{n} = {a}^{mn}}}[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{3 \times \frac{1}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{5 \times \frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{\frac{3}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{\frac{5}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times\Bigg[\bigg(\frac{3}{2} \bigg)^{1} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \times \dfrac{3}{2} \bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3 \times 3}{2 \times 2}\bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]

[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)\times \Bigg[\bigg(\frac{3}{2} \bigg)\times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3}{2} \times \dfrac{9}{4} \: \: \Bigg]}}\\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3 \times 9}{2 \times 4} \: \: \Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg(\dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{27}{8} \: \: \Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\dfrac{3}{2} \times \dfrac{27}{8}}} \\[/tex]

[tex]\sf{\longmapsto{\dfrac{3 \times 27}{2 \times 8}}} \\[/tex]

[tex] \sf{\longmapsto{\dfrac{81}{16}}\: ≈ \:5.0625\:\red{Ans.}} \\[/tex]

Answer:

27s

Step-by-step explanation:

d = distance to rendezvous point in miles

t = time of arrival at 10mph, i.e. 3 mins later than appointed time

10 mph = 10 miles per hour = ¹/₆ miles per min

12 mph = ¹/₅ miles per min

t = d/(¹/₆)

t = 6d

5 min = ¹/₁₂ hr

t - ¹/₁₂ = d/(¹/₅)

t = 5d + ¹/₁₂

Elimination of t:

6d = 5d + ¹/₁₂

d = ¹/₁₂ mi

t = 6(¹/₁₂)

t = ¹/₂ min

Time left to appointed time = ¹/₂ - ³/₆₀

= ¹⁰/₂₀ - ¹/₂₀

= ⁹/₂₀.min → 27 seconds

Answer:

27 min

Step-by-step explanation:

rsm

Answer:

Step-by-step explanation:

We know that:

We observe the denominator of the first expression is same as the product of the denominators of the second expression.

Compare the numerators and solve for b:

[tex](4a - 5b)^{2} \\ by \: \: \: using \: \: \: (x - y)^{2} = {x}^{2} - 2xy + {y}^{2} \\ = {(4a)}^{2} - 2(4a)(5b) + {(5b)}^{2} \\ = {16a}^{2} - 40ab + 25 {b}^{2} [/tex]

[tex] {16a}^{2} - 40ab + {25b}^{2} [/tex]

Hope it helps.

Do comment if you have any query.

Answer:

x=g-c

Step-by-step explanation:

Answer:

no

yes

yes

no

no

yes

Step-by-step explanation:

a prime number cant be devided by any thing other than itself or one

Answer:

Step-by-step explanation:

Is 2 is 3 is 11 is 18 is 20 is 27

Answer:

y=[tex]-\frac{3}{4}[/tex]x+2

Step-by-step explanation:

y=sx+2;

-1=4s+2;

s=-3/4;

y=[tex]-\frac{3}{4}[/tex]x+2

Answer: The second bubble thing is correct you have picked the wrong one.