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3 years ago

A bathtub is being filled up and after 2 3/4 minutes there

1 answer:

8 0

Well, to solve this problem, we can simply divide 8 and 1/2 by 2 and 3/4 so, let’s do that.

(8 1/2) / (2 3/4)
= (17/2) / (11/4). ***dividing two fractions together is the same thing as multiplying them but the second fraction is flipped.

(17/2) / (11/4)
= (17/4) * (4/11)
= 68/44

So, in conclusion, the answer to this question is: 68/44.
=

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PLEASE SOLVE QUESTION 13 AND 14 DONT STEAL THE POINTS

Varvara68 [4.7K]

Answer:

See below for solution.

Step-by-step explanation:

$Q13)$ $\frac{\sqrt{11} - \sqrt{5} }{\sqrt{11}+\sqrt{5} }$ = $x - y\sqrt{55}$

LHS :-

= $\frac{\sqrt{11} - \sqrt{5} (\sqrt{11}-\sqrt{5}) }{\sqrt{11}+\sqrt{5} (\sqrt{11}-\sqrt{5}) }$

= $\frac{11-2\sqrt{55}+5 }{11 - 5}$

= $\frac{16-2\sqrt{55} }{6}$

= $\frac{2(8-\sqrt{55}) }{2(3)}$

= $\frac{8-\sqrt{55} }{3}$

=> x = 8/3 and y = 1/3

=> Option B

$Q14)$ $\frac{4}{216^{-2/3} }$ + $\frac{1}{256^{-3/4} }$ + $\frac{2}{243^{-1/5} }$

= 4 x (∛216)² + ( $\sqrt[4]256}$ )³ + 2 x ( $\sqrt[5]{243}$ )

= 4 x 36 + 64 + 2 x 3

= 144 + 64 +6

= 144 + 70

= 214

3 0

2 years ago

Setler [38]

1 is A the domain is the X values meanwhile y is your range.

4 0

4 years ago

Write an equation and find missing angle . 6x ° and 30 °

s2008m [1.1K]

6x+30=180

6×=180-30

×=150/6

×=25

6x=6×25=130

5 0

3 years ago

ANSWER ASAP !! NEED HELP ASAP

liq [111]

Answer:

130

Step-by-step explanation:

The two angles add up to be 180 degrees, so you could find the value of the second angle to be 110 degrees, but it is asking for the x value so you would choose 130 since when you subtract 20 from 130 you get 110.

4 0

3 years ago

Given: ΔTSP, TS = SP = 10cm, TP = 12cm. Find: Three altitudes of ΔTSP.

fredd [130]

Answer:

The altitudes of ΔTSP is SC is 8 unit ,

PA is 5 $\sqrt{3}$ unit ,

TB is 5 $\sqrt{3}$ unit .

Step-by-step explanation:

Given as :

The Triangle Δ TSP having side TS , SP , TP

The measure of TS = 12 cm

The measure of SP = 10 cm

The measure of TP = 10 cm

Let The altitude from point T on side PS = TB

The altitude from point S on side PT = SC

The altitude from point P on side TS = PA

The altitude divide the sides

TS = TA + AS

PS = PB + PS

PT = PC + CT

So, From Pythagoras theorem

PA² = PS² - AS²

PA² = 10² - 5²

Or, PA² = 100 - 25

Or, PA = $\sqrt{75}$

∴ PA = 5 $\sqrt{3}$ unit

Again

TB² = TS² - BS²

TB² = 10² - 5²

Or, TB² = 100 - 25

Or, TB = $\sqrt{75}$

∴ TB = 5 $\sqrt{3}$ unit

Similarly

SC² = TS² - TC²

SC² = 10² - 6²

Or, SC² = 100 - 36

Or, SC = $\sqrt{64}$

∴ SC = 8 unit

Hence The altitudes of ΔTSP is SC is 8 unit , PA is 5 $\sqrt{3}$ unit , TB is 5 $\sqrt{3}$ unit . Answer

4 0

4 years ago