Help me with questions 3 and 4 please​!

Answer:

|-4| < |-5|

Step-by-step explanation:

because if modules is given sub sign will be deducate

Answer:

im not sure

Step-by-step explanation:

Answer:

95m

Step-by-step explanation:

312 - 61 - 61 = 190

190/2 = 95

Answer:

x=g-c

Step-by-step explanation:

9514 1404 393

Answer:

  63

Step-by-step explanation:

The number of seats in a row will give an arithmetic sequence:

  6, 9, 12, 15, ...

The first term is 6; the common difference is 3. The general term is ...

  an = a1 +d(n -1) . . . . . . n-th term of sequence with first term a1, difference d

The 20th term of the sequence is ...

  a20 = 6 +3(20 -1) = 6 +57 = 63

There would be 63 seats on the 20th row.

Answer:

.......................

Step-by-step explanation:

= f( g(2) )

= 3(2x + 1) + 1

= 6x + 3 + 1

= 6x + 4

= 6(2) + 4

= 12 + 4

= 16

f(g(2)) = 16

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

Answer:

20

Step-by-step explanation:

20*20 = 400

OR

[tex]20 {}^{2} = 400[/tex]

Answer:

9 hours worked after school and 4 hours worked on saturday

Step-by-step explanation:

x = amount of hours worked after school

y = amount of hours worked on saturday

x + y = 13 or x = 13 - y

and

3x + 4y = 43

plug:

3 · (13 - y) + 4y = 43

39 - 3y + 4y = 43

y = 4

plug:

x = 13 - y

x = 13 - 4 = 9

9 hours worked after school and 4 hours worked on saturday

Answer: i solved on my channel

Step-by-step explanation: https://youtu.be/rTgj1HxmUbg

Answer:

makes it slope

Step-by-step explanation:

Answer: 9

Step-by-step explanation:

36 ÷ 4 = 9

Answer:

x = 49

Step-by-step explanation:

Since the triangles are similar then the ratios of corresponding sides are in proportion, that is

[tex]\frac{DN}{KJ}[/tex] = [tex]\frac{DG}{KM}[/tex] , substitute values

[tex]\frac{5}{35}[/tex] = [tex]\frac{8}{x+7}[/tex] ( cross- multiply )

5(x + 7) = 280 ( divide both sides by 5 )

x + 7 = 56 ( subtract 7 from both sides )

x = 49

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:

a) 2 * 3 = 6.

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

Step-by-step explanation:

Step-by-step explanation:

cost per day

= $1500 / week

= $1500 / 7 days

= $1500 ÷ 7 / 7 ÷ 7

≈ $214,29 / day

Answer:

z

Step-by-step explanation:

hqvqhqvw karrar ras wallah a part

Answer:

3/4

Step-by-step explanation:

simply divide 3 1\4‐2 1/2

question ;

Pinky had 3 ¼ metre of red ribbon. She 2 ½ used metre of ribbon to wrap gifts. How much ribbon is left with her ?

answer ;

Answer:

Step-by-step explanation:

To find the y-intercept, set x=0 and evaluate the function.

  f(0) = -3/(0 -4) = 3/4

The y-intercept is (0, 3/4).

__

To find the x-intercept(s), set f(x) = 0 and solve for x.

  0 = -3/(x^2 -4)

  0 = -3 . . . . . . . . . . multiply by (x^2 -4), x ≠ ±2

There are no values of x that will make this true. There are no x-intercepts.

_____

Additional comments

In general, you find the x-intercepts of a rational function by finding the zeros of the numerator. Here, the numerator is -3, so cannot ever be zero.

I find a graphing calculator to be a useful tool for showing where to look for x-and y-intercepts. The attached graph shows y=0 (the x-axis) is a horizontal asymptote, so there are no x-intercepts.

Answer:

Step-by-step explanation:

[tex]2\sqrt{8}+3\sqrt{50}=2\sqrt{2*2*2} + 3\sqrt{5*5*2}\\\\= 2*2\sqrt{2}+3*5\sqrt{2}\\\\=4\sqrt{2}+15\sqrt{2}\\\\= 19\sqrt{2}[/tex]

[tex]3\sqrt{12}+4\sqrt{27}=3\sqrt{2*2*3}+4\sqrt{3*3*3}\\\\=3*2\sqrt{3}+4*3\sqrt{3}\\\\=6\sqrt{3}+12\sqrt{3}\\\\=(6+12)\sqrt{3}\\\\= 18\sqrt{3}[/tex]

Answer:

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

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]

[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:

yes

7164 is divisible bt 6

Answer:

Yes

Step-by-step explanation:

Any number will be divisible by 6 if they be divisible by 2 & 3

- we know 7164 is divisible by 2

- also any number which sum of their digits are divisible by 3, the number will be divisible by 3

in this case (7164) 7+1+6+4=18 & 18 is divisible by 3 so 7164 is divisible by 3 as well.

so 7164 is divisible by 6

Answer:

[tex]\frac{\sqrt{70} }{5}[/tex]