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babunello [35]
3 years ago
14

What is the value of x? Enter your answer in the box. x = ​​

Mathematics
1 answer:
nexus9112 [7]3 years ago
3 0

Answer: x= 18.2

Step-by-step explanation:

We can assume that this is a right triangle meaning it is 90 degrees, so

2x-1 + 3x must = to 90 degrees which gives us,

2x-1+3x= 90 => 2x+3x-1=90 (Rearrange like terms) => 5x-1=90 => 5x=91 => x= 18.2

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ANSWER SHEET
faust18 [17]

Answer:

.

Step-by-step explanation:

5 0
3 years ago
Given that h(x) = -3x - 7, what is h(-4)?
iris [78.8K]

Answer:

x=−3x−7whatish(−4)

Step 1: Flip the equation.

28ah2istw−3x=x

Step 2: Add 3x to both sides.

28ah2istw−3x+3x=x+3x

28ah2istw=4x

Step 3: Divide both sides by 28h^2istw.

28ah2istw

28h2istw

=

4x

28h2istw

a=

x

7h2istw

Answer:

a=x7h2istw

3 0
3 years ago
Read 2 more answers
If a box has a surface area of 680 and the length and width are 10. What is the height?
Molodets [167]

10x10=100

680/100=6.8

10x10x6.8=680

the answer is 6.8

8 0
4 years ago
Solve 2x - 11 = k for x.
denis23 [38]
2x + -11 = k

Reorder the terms:
-11 + 2x = k

Solving
-11 + 2x = k

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '11' to each side of the equation.
-11 + 11 + 2x = 11 + k

Combine like terms: -11 + 11 = 0
0 + 2x = 11 + k
2x = 11 + k

Divide each side by '2'.
x = 5.5 + 0.5k

Simplifying
3 0
4 years ago
Sea un cuadrado de 2 pulgadas de lado uniendo los puntos medios se obtiene otro cuadrado inscrito en el anterior si repetimos es
Ne4ueva [31]

Answer:

1) La serie geométrica formada es

4, 2, 1,..., ∞

2) La suma al infinito de las áreas de los cuadrados es 8 in.²

Step-by-step explanation:

1) El área del primer cuadrado, a₁ = 2² = 4 pulgadas²

El área del siguiente cuadrado, a₂ = (√ (1² + 1²)) ² = (√2) ² = 2 pulg²

El área del siguiente cuadrado, a₃ = ((√ (2) / 2) ² + (√ (2) / 2) ²) = 1 pulg²

Por lo tanto, la razón común, r = a₂ / a₁ = 2/4 = a₃ / a₂ = 1/2

Las áreas de los cuadrados progresivos forman una progresión geométrica como sigue;

4, 4×(1/2), 4 ×(1/2)²,...,4×(1/2)^{\infty}

De donde obtenemos la serie geométrica formada de la siguiente manera;

4, 2, 1,..., ∞

2) La suma de 'n' términos de una progresión geométrica hasta el infinito para -1 <r <1 se da como sigue;

S_{\infty} = \dfrac{a}{1 - r}

Por lo tanto, la suma de las áreas de los cuadrados hasta el infinito se obtiene sustituyendo los valores de 'a' y 'r' en la ecuación anterior de la siguiente manera;

La \ suma \ al \ infinito \ del \ cuadrado \ S_{\infty}  = \dfrac{4 \ in.^2}{1 - \dfrac{1}{2} } = \dfrac{4 \ in.^2}{\left(\dfrac{1}{2} \right)} = 2 \times 4 \ in.^2= 8 \ in.^2

La suma al infinito de las áreas de los cuadrados, S_{\infty} = 8 in.²

7 0
3 years ago
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