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adelina 88 [10]

3 years ago

9

Josiah had 1/5 + 1/2 he Sinister was 2/7

1 answer:

miskamm [114]3 years ago

5 0

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- 3(3s - 4) - 12 = - 3/6s + 1) - 4

s344n2d4d5 [400]

Answer:

s=-7/9

Step-by-step explanation:

-9s+12-12=-18s-3-4

-9s=-18s-7

9s=-7

s=-7/9

(if I understood the question correctly)

6 0

3 years ago

There are 24 students in class. 75% completed their homework early. How many students completed their homework early?

andrezito [222]

16 students turned in their homework early i think. I hope this helps

5 0

3 years ago

A 2000 L tank containing 550 L of water is being filled with water at the rate of 75 L per minute from a full 1600 L tank. How l

lara [203]

Answer:

After 7 minutes both the tanks will have the same amount of water.

Step-by-step explanation:

The water is filling in a tank at the rate of 75 L/mins from an initial 550 L of water and water is removing from another tank at the same rate from the initial 1600 L of water.

Let us assume that the same amount of water will be there in both the tanks after t minutes.

Therefore, 550 + 75t = 1600 - 75t

⇒ 150t = 1050

⇒ t = 7

Therefore, after 7 minutes both the tanks will have the same amount of water. (Answer)

7 0

3 years ago

Hello , anyone able to help with question 5? thanks :)!

astra-53 [7]

Answer:

see explanation

Step-by-step explanation:

Using the product/ difference identity

sinxsiny = $\frac{1}{2}$ [ cos(x - y) - cos(x + y) ]

Consider the left side

2sin( $\frac{\pi }{4}$ + α)sin( $\frac{\pi }{4}$ - α )

= cos[ $\frac{\pi }{4}$ + α - $\frac{\pi }{4}$ + α ] - cos [ $\frac{\pi }{4}$ + α + $\frac{\pi }{4}$ - α ]

= cos2α - cos $\frac{\pi }{2}$

= cos2α - 0

= cos2α = right side , thus proven

4 0

2 years ago

Please help if you know !! Much appreciated

kozerog [31]

QUESTION 1

Recall the mnemonics; SOH

$\sin L =\frac{Opposite}{Hypotenuse}$

$\sin L =\frac{12}{13}$

QUESTION 2

Recall the mnemonics; CAH

$\cos L =\frac{Adjacent}{Hypotenuse}$

$\sin L =\frac{5}{13}$

QUESTION 3

Recall the mnemonics; TOA

$\tan M =\frac{Opposite}{Adjacent}$

$\tan M =\frac{5}{12}$

QUESTION 4

$\sin M =\frac{Opposite}{Hypotenuse}$

$\sin M =\frac{5}{13}$

QUESTION 5

a) From the Pythagoras Theorem,

$AC^2+AB^2=BC^2$

$AC^2+4^2=5^2$

$AC^2+16=25$

$AC^2=25-16$

$AC^2=9$

$AC=\sqrt{9}$

$AC=3yd$

b) Using the cosine ratio,

$\cos (m\angle B)=\frac{4}{5}$

Take the inverse cosine of both sides;

$m\angle B=\cos ^{-1}(\frac{4}{5})$

$m\angle B=36.86$

$\therefore m\angle B=36.9\degree$ to the nearest tenth.

c) $m\angle B +m\angle C=90\degree$

$\Rightarrow 36.9\degree +m\angle C=90\degree$

$\Rightarrow m\angle C=90\degree-36.9\degree$

$\Rightarrow m\angle C=53.1\degree$

QUESTION 6

a) using the sine ratio,

$\sin(51\degree)=\frac{DE}{18}$

$DE=18\sin(51\degree)$

$DE=13.99$

$\therefore DE=14.0yd$ to the nearest tenth.

b) Using the cosine ratio,

$\cos (51\degree)=\frac{EF}{18}$

$EF=18\cos (51\degree)$

$EF=11.3yd$ to the nearest tenth.

c) $m\angle D+ m\angle F=90\degree$

$\Rightarrow m\angle D + 51\degree=90\degree$

$\Rightarrow m\angle D=90\degree-51\degree$

$\Rightarrow m\angle D=39\degree$

QUESTION 7

a) Using the Pythagoras Theorem;

$lG^2+GH^2=lH^2$

$l15^2+GH^2=17^2$

$225+GH^2=289$

$GH^2=289-225$

$GH^2=64$

$GH=\sqrt{64}$

$GH=8km$

b) Using the sine ratio,

$\sin(m\angle H)=\frac{15}{17}$

$m\angle H=\sin^{-1}(\frac{15}{17})$

$m\angle H=61.9\degree$

c) $m\angle l + m\angle H=90\degree$

$m\angle l +61.9\degree=90\degree$

$m\angle l =90\degree-61.9\degree$

$m\angle l =28.1\degree$

QUESTION 8

We plot the points as shown in the diagram.

QUESTION 9

From the diagram, the side lengths XY and YZ can be obtained by counting the boxes. Each box is 1 unit.

This implies that;

XY =3 units

YZ=5 units.

We use Pythagoras Theorem, to obtain XZ.

This implies that;

$XZ^2=XY^2+YZ^2$

$XZ^2=3^2+5^2$

$XZ^2=9+25$

$XZ^2=34$

$XZ=\sqrt{34}$ units

QUESTION 10.

a) Using the tangent ratio;

$\tan(m\angle X)=\frac{5}{3}$

$m\angle X=\tan^[-1}(\frac{5}{3})$

$m\angle X=59.0\degree$

b) $m\angle Z+m\angle X=90\degree$

$m\angle Z+59.0\degree=90\degree$

$m\angle Z=90\degree-59.0\degree$

$m\angle Z=31.0\degree$

QUESTION 11

a) Triangle BCD is shown in the attachment.

The length of side DC=|3-2|=1 unit

The length of side DB=|4-3|=1 unit

Using Pythagoras Theorem;

$BC^2=DC^2+DB^2$

$BC^2=1^2+1^2$

$BC^2=1+1$

$BC^2=2$

$BC=\sqrt{2}$

b) DB is perpendicular to DC, therefore m<D=90 degrees.

The length of DB is equal the length of DC.

This implies that;

m<C=m<B=45 degrees.

QUESTION 12

$\sin 30\degree=\frac{1}{2}$

QUESTION 13

$\cos 30\degree=\frac{\sqrt{3} }{2}$

QUESTION 14

$\tan 30\degree =\frac{\sqrt{3} }{3}$

QUESTION 15

$\tan 45\degree =1$

QUESTION 16

$\cos 45\degree =\frac{\sqrt{2} }{2}$

QUESTION 17

$\tan 45\degree =1$

<h3>Check attachment for the rest</h3>

3 0

3 years ago