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vagabundo [1.1K]
3 years ago
14

Pls help me!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!please I’m begging u

Mathematics
2 answers:
Natasha2012 [34]3 years ago
6 0

Answer:

7. 36     8. 1     9. 21

Step-by-step explanation:

You plug in the variables in the equation.

7.5^{2} +-4^{2} = 41

8. (5-4)^{2} = 81

9. 5+4^{2} =21

lions [1.4K]3 years ago
5 0

Answer:

See explanation

Step-by-step explanation:

1. 8^5 = 32768

2. a^6

3. 5^5*9^3 = 2278125

5. (-3)^5 = -243

6. (3/4)³ = 27/64

7. a² + b² = 5² + (-4)² = 25 + 16 = 41

8. (a + b)² = (5 + 4)² = 9² = 81

9. a + b² = 5 + 4² = 5 + 16 = 21

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Working alone, Ryan can dig a 10 ft by 10 ft hole in 5 hours Castel can dig the same hole in 6 hours how long would it take them
Fed [463]

Answer:

The time taken by both boys to complete work = 2.73 hours

Step-by-step explanation:

Time taken by rayan to dig hole  T_{1} = 5 hours

Time taken by castel to dig hole T_{2} = 6 hours

Now the time taken by both persons working together  

T = \frac{T_{1} T_{2}  }{T_{1} + T_{2}}

Put all the values in above formula we get

T = \frac{ 30 }{6 + 5}

T = \frac{30}{11}

T = 2.73 hours

Therefore the time taken by both boys to complete work = 2.73 hours

6 0
3 years ago
La siguiente tabla presenta las frecuencias absolutas y relativas de las distintas caras de un dado cuando se simulan 300 lanzam
noname [10]

La opinión correcta es la de Camila que argumenta que la frecuencia relativa de la cara con el #6 será un valor cercano a 0,166.

<h3>Qué es la frecuencia relativa?</h3>

La frecuencia relativa es un término matemático que se utiliza en la estadística para referirse al número de veces que un evento se repite durante un experimento.

Por otra parte, es necesario aclara que la frecuencia relativa no se modifica en gran medida si se aumenta el número de veces de una prueba, debido a que la probabilidad de ocurrencia de este evento va a ser la misma.

De acuerdo a lo anterior, Camila tiene razón debido a que considera que el valor de la frecuencia relativa no se modifica en gran medida entre 300 o 600 lanzamientos de los dados.

Nota: Esta pregunta está incompleta porque no está la tabla. No obstante, la puedo responder basado en mi conocimiento previo.

Aprenda más sobre probabilidad en: brainly.com/question/16019390

6 0
2 years ago
Use implicit differentiation to find y^1 for the equation y^2-y-4x=0
Rudik [331]

Answer:

Step-by-step explanation:

y²-y-4x=0

differentiate w.r.t. x

2y*y^1-y^1-4=0

(2y-1)y^1=4

y^1=4/(2y-1)

3 0
3 years ago
The subject of the formula below is y.<br>a=4x/t - p<br>Rearrange the f make x the subject.​
Crazy boy [7]

Answer:

x = t(a+p)/4

Step-by-step explanation

Given the expression

​a=4x/t - p

We are to make x the subject of the formula

a=4x/t - p

Add p to both sides

a+p = 4x/t - p+p

a+p = 4x/t

Cross multiply

t(a+p) = 4x

Rearrange

4x = t(a+p)

Divide both sides by 4

4x/4 = t(a+p)/4

x = t(a+p)/4

4 0
3 years ago
Solve.
Nikitich [7]
The answer is C: 10\text{ }^1/_2. Here are the details:

\text{Equation:}\\ 6\text{ }^1/_3+10\text{ }^1/_2\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{Start with the integers.}\\ 6+10=16\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{...then with the fractions, but rewrite them first to make it easier.}\\ ^1/_3+\text{ }^1/_2\stackrel{\text{rewrite}}{\to}\text{ }^2/_6+\text{ }^3/_6=\text{ }^5/_6\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{Add 'em up!}\\&#10;16\text{ }^5/_6

\text{Equation:}\\&#10;16\text{ }^5/_6+3\text{ }^5/_6\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{Start with the integers.}\\&#10;16+3=19\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{...then with the fractions, but rewrite them first to make it easier.}\\&#10;^5/_6+\text{ }^5/_6=\text{ }^{10}/_6\stackrel{\text{rewrite}}{\to}\text{ }^5/_3\stackrel{\text{rewrite}}{\to}1\text{ }^2/_3\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{Add 'em up!}\\&#10;20\text{ }^2/_3

\text{Last equation:}\\ 20\text{ }^2/_3+5\text{ }^1/_2\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{Start with the integers.}\\ 20+5=25\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{...then with the fractions, but rewrite them first to make it easier.}\\ ^2/_3+\text{ }^1/_2\stackrel{\text{rewrite}}{\to}\text{ }^4/_6+\text{ }^3/_6=\text{ }^7/_6\stackrel{\text{rewrite}}{\to}1\text{ }^1/_6\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{Add the integer and fraction together.}\\&#10;25+1\text{ }^1/_6\stackrel{?}{=}26\text{ }^1/_6

26\text{ }^1/_6\stackrel{\checkmark}{=}26\text{ }^1/_6
5 0
3 years ago
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