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ankoles [38]
2 years ago
7

Just do numbers 1-6,22-24, and 25-26

Mathematics
1 answer:
Paul [167]2 years ago
7 0

1. 0.625

2. 0.22222222222

3. 0.43243243243

4. -0.11111

5. 0.54

6. -0.75

22. −2 21/50

23. -4/25

24. -13/20

25. 0.62

26. 7/15

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What is the following quotient? sqrt6+sqrt11/sqrt5+sqrt3​
Gnoma [55]

Answer:

√30-√18+√55-√33/2

Step-by-step explanation:

What is the following quotient? sqrt6+sqrt11/sqrt5+sqrt3​

Given the expression

√6+√11/√5+√3

Rationalize

√6+√11/√5+√3 × √5-√3/√5-√3

= √30-√18+√55-√33/√25-√9

= √30-√18+√55-√33 /5-3

= √30-√18+√55-√33/2

This gives the required quotient

6 0
3 years ago
I was taught how to do these I just don’t remember how to-3(4x+5)-2(-5x+8)
frosja888 [35]

Answer:

- 2x - 31

Step-by-step explanation:

Given

- 3(4x + 5) - 2(- 5x + 8)

Multiply the terms in the first parenthesis by - 3 and the terms in the second parenthesis by - 2

= - 12x - 15 + 10x - 16 ← collect like terms

= (- 12x + 10x) + (- 15 - 16)

= - 2x - 31

6 0
3 years ago
Six to the third power times six to the fourth power in exponential form
Marta_Voda [28]

Answer:

6^7

Step-by-step explanation:

<u>Step 1:  Convert</u>

Six to the third power times six to the fourth power.

6^3 * 6^4

6^(^3^+^4^)

6^7

Answer:  6^7

3 0
3 years ago
Read 2 more answers
10. Given that the total distance travelled is 4 km and that the initial aceleration is 4 m/s², find:a) the value of V.b) the va
andrezito [222]

The initial deceleration is 4 m/s^2, so this means that considering the first line segment, \frac{V-100}{T}=-4 \implies V=100-4T.

The total distance traveled is 4 km, which is equal to 4000 meters. This is equal to the area under the graph, so 400T-\frac{1}{2}(100-V)(5T)=4000.

400(100-4T)-\frac{1}{2}(4T)(5T)=4000\\\\40000-1600T-10T^2=4000\\\\10T^2 +1600T-36000=0\\\\T^2 +160T-3600=0\\\\(T+180)(T-20)=0\\\\T=20 (T > 0)\\\\\therefore V=100-4(20)=20

6 0
1 year ago
Triangles P Q R and S T U are shown. Angles P R Q and T S U are right angles. The length of P Q is 20, the length of Q R is 16,
dimaraw [331]

Answer:

\angle P

Step-by-step explanation:

Given

\triangle PRQ = \triangle TSU = 90^o

PQ = 20     QR = 16    PR = 12

ST = 30       TU = 34    SU = 16

<em>See attachment</em>

Required

Which sine of angle is equivalent to \frac{4}{5}

Considering \triangle PQR

We have:

\sin(P) = \frac{QR}{PQ} --- i.e. opposite/hypotenuse

So, we have:

\sin(P) = \frac{16}{20}

Divide by 4

\sin(P) = \frac{4}{5}

Hence:

\angle P is correct

7 0
2 years ago
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