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3 years ago

9

ABCD is a rectangle. Find the length of each diagonal...

1 answer:

Dominik [7]3 years ago

4 0

Setting AC = BD because the diagonals are congruent, then 3y/5 = 3y -4
y=5/3
So then AC = 1 = BD

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Angelo thinks of a number, multiplies it by 5 and subtracts 8.His number is 47,wite as an equation in y and solve to find Angelo

Inessa05 [86]

Answer:

<h3>y*5-8=47</h3><h3>y*5=47+8</h3><h3>y*5=55</h3><h3>y=55/5</h3><h3>y=11</h3>

7 0

3 years ago

En un triángulo rectángulo A es un ángulo agudo y Sen A = 4/5 ¿Cuál será el valor de Tan A?

Nonamiya [84]

Answer:

$\displaystyle \tan A=\frac{4}{3}$

Step-by-step explanation:

<u>Funciones Trigonométricas</u>

La identidad principal en trigonometría es:

$sen^2A+cos^2A=1$

Si sabemos que A es un ángulo agudo (que mide menos de 90°), su seno y coseno son positivos.

Dado que Sen A = 4/5, calculamos el coseno:

$cos^2A=1-sen^2A$

Sustituyendo:

$\displaystyle cos^2A=1-\left(\frac{4}{5}\right)^2$

$\displaystyle cos^2A=1-\frac{16}{25}$

$\displaystyle cos^2A=\frac{25-16}{25}$

$\displaystyle cos^2A=\frac{9}{25}$

Tomando raíz cuadrada:

$\displaystyle cos\ A=\sqrt{\frac{9}{25}}=\frac{3}{5}$

La tangente se define como:

$\displaystyle \tan A=\frac{sen\ A}{cos\ A}$

Substituyendo:

$\displaystyle \tan A=\frac{\frac{4}{5}}{\frac{3}{5}}$

$\displaystyle \tan A=\frac{4}{3}$

6 0

3 years ago

In the next olympics the united states can enter four athletes in the divivng competition. how many different teams of four dive

n200080 [17]

If you have 9 divers total and each team can only have 4 divers, then logically the US can select 2 teams of 4 from 9 divers.

4 0

3 years ago

Identify the slope and y-intercept of the graph of the equation.

Ksivusya [100]

Answer:

B

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Pls solve the simultaneous equation in the attachment.

siniylev [52]

Answer:

Part a) The solution is the ordered pair (6,10)

Part b) The solutions are the ordered pairs (7,3) and (15,1.4)

Step-by-step explanation:

Part a) we have

$\frac{x}{2}-\frac{y}{5}=1$ ----> equation A

$y-\frac{x}{3}=8$ ----> equation B

Multiply equation A by 10 both sides to remove the fractions

$5x-2y=10$ ----> equation C

isolate the variable y in equation B

$y=\frac{x}{3}+8$ ----> equation D

we have the system of equations

$5x-2y=10$ ----> equation C

$y=\frac{x}{3}+8$ ----> equation D

Solve the system by substitution

substitute equation D in equation C

$5x-2(\frac{x}{3}+8)=10$

solve for x

$5x-\frac{2x}{3}-16=10$

Multiply by 3 both sides

$15x-2x-48=30$

$15x-2x=48+30$

Combine like terms

$13x=78$

$x=6$

<em>Find the value of y</em>

$y=\frac{x}{3}+8$

$y=\frac{6}{3}+8$

$y=10$

The solution is the ordered pair (6,10)

Part b) we have

$xy=21$ ---> equation A

$x+5y=22$ ----> equation B

isolate the variable x in the equation B

$x=22-5y$ ----> equation C

substitute equation C in equation A

$(22-5y)y=21$

solve for y

$22y-5y^2=21$

$5y^2-22y+21=0$

Solve the quadratic equation by graphing

The solutions are y=1.4, y=3

see the attached figure

<em>Find the values of x</em>

For y=1.4

$x=22-5(1.4)=15$

For y=3

$x=22-5(3)=7$

therefore

The solutions are the ordered pairs (7,3) and (15,1.4)

3 0

3 years ago