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3 years ago

14

Please help with questions a, b and c!!

1 answer:

Makovka662 [10]3 years ago

4 0

A) y=1/2 + 5

B) y=-1/2 + 8

C) y=1/2 - 10

Slope = M = Gradient

y-intercept = y-value = on y-coordinate

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NEED HELP WORTH 100 POINTS SHOW YOUR WORK PLEASE

tia_tia [17]

Answer:

Q1:

$-5\sqrt{27c^2}$

$-5\sqrt{27}\sqrt{c^2}$

$-5\times\sqrt{9} \sqrt{3} \sqrt{c^2}$

$-5\times3\sqrt{3}\sqrt{c^2}$

$-5\times3\sqrt{3}c$

$-15\sqrt{3}c$

Q2:

$5\sqrt{72}-3\sqrt{32}$

$5\sqrt{36} \sqrt{2} -3\sqrt{16} \sqrt{2}$

$5\times6\sqrt{2}-3\times4\sqrt{2}$

$30\sqrt{2}-12\sqrt{2}$

$18\sqrt{2}$

Q3:

$\sqrt{108yz}+3\sqrt{98yz}+2\sqrt{75yz}$

$\sqrt{36} \sqrt{3} \sqrt{yz} +3\sqrt{49} \sqrt{2} \sqrt{yz} +2\sqrt{25} \sqrt{3} \sqrt{yz}$

$6\sqrt{3}\sqrt{yz}+21\sqrt{2}\sqrt{yz}+10\sqrt{3}\sqrt{yz}$

$16\sqrt{3}\sqrt{yz}+21\sqrt{2}\sqrt{yz}$

Q4:

$\sqrt{15n^2}\sqrt{10n^3}$

$\sqrt{15}n\sqrt{10n^3}$

$\sqrt{15}n\sqrt{10}\sqrt{n^2}\sqrt{n}$

$\sqrt{15}\sqrt{10}n^2\sqrt{n}$

$\sqrt{5}\sqrt{3}\sqrt{5}\sqrt{2}n^2\sqrt{n}$

$5\sqrt{3}\sqrt{2}n^2\sqrt{n}$

$5\sqrt{6}n^2\sqrt{n}$

Q5:

$\frac{\sqrt{3x^2y^3}}{4\sqrt{5xy^3}}$

$\frac{\sqrt{3}xy\sqrt{y}}{4\sqrt{5}y\sqrt{x}\sqrt{y}}$

Cancel off $\sqrt{x}$ :

$\frac{\sqrt{3}y\sqrt{x}\sqrt{y}}{4\sqrt{5}y\sqrt{y}}$

Cancel off $y:$

$\frac{\sqrt{3}\sqrt{x}\sqrt{y}}{4\sqrt{5}\sqrt{y}}$

Cancel off $\sqrt{y} :$

$\frac{\sqrt{3}\sqrt{x}}{4\sqrt{5}}$

Multiply $\sqrt{5}$ on the numerator and denominator:

$\frac{\sqrt{3}\sqrt{x}\sqrt{5}}{4\sqrt{5}\sqrt{5}}$

$\frac{\sqrt{15}\sqrt{x}}{20}$

5 0

3 years ago

Twenty-one boys and eight girls go on a camping trip. In how many ways can a group of six be selected to gather firewood, given

lorasvet [3.4K]

(a) there are 8C2 = 28 ways of picking 2 girls from 8
And there are 21C4 = 5985 ways of picking 4 boys

Required number of ways for 2g / 4b = 28 * 5985 = 167,580

(b) at least 2 girls means combinations of 2g/4b , 3g,3b , 4g/2b , 5g 1b or
6 girls.

2g/4b = 167,580 ways
3g/3b = 8C3 * 21C3 = 56 * 1330 = 74,480
4g/2b = 8C4* 21C2 = 70 * 210 = 14,700
5g 1b = 8C5* 21 = 56*21 = 1176
6 girls = 8C6 = 28

adding these up we get the answer to (b) which is 257,964

6 0

3 years ago

Select all the expressions equivalent to: (4^5)⋅4^3 Group of answer choice 2^14 4^12 2^16 16^8 4^8 16^4

kumpel [21]

Answer:

Choices given as $2^{16}, 4^8$ and $16^4$ are correct.

Step-by-step explanation:

Given expression is,

$(4^5)(4^3)$

By applying law of exponents in the given expression,

$(4^5)(4^3)=4^{5+3}$

$=4^8$

$(4^5)(4^3)=(2^2)^5(2^2)^3$

$=(2^{2\times 5})(2^{2\times 3})$

$=2^{10}\times 2^6$

$=2^{16}$

$=(2^4)^4$

$=(16)^4$

Therefore, choices given as $2^{16}, 4^8$ and $16^4$ are correct.

4 0

3 years ago

Can 6 fit into 36 and if so how many times

ratelena [41]

Yes.
This is the same thing as:
$\frac{36}{6} =6$

How many times can 6 go into 36?
6 times
Try:
6*6=36

So true.

Hope this helps! :D

6 0

3 years ago

Read 2 more answers

Solve the equation on the interval [0,2π). 7 sec x-7=0, x=?

Oksanka [162]

$\huge \bf༆ Answer ༄$

Let's solve ~

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:7 \sec(x) - 7 = 0$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:7 \sec(x) = 7$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \: \sec(x) = 7 \div 7$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \: \sec(x) = 1$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:x = 0$

8 0

2 years ago