What two numbers satisfy the following two conditions? -Multiplies to -75 -Adds to 10

In the isosceles △ABC m∠ACB=120° and AD is an altitude to leg BC . What is the distance from D to base AB , if CD=4cm?

snow_lady [41]

If ΔACB is an isosceles triangle, then ∠A ≅ ∠B and AC ≅ CB

Since ∠C = 120° and ∠A + ∠B + ∠C = 180°, then ∠A = 30° and ∠B = 30°

Next, look at ΔADB. ∠A + ∠D + ∠B = 180°, so ∠A + 90° + 30° = 180° ⇒ ∠A = 30°

Now look at ΔADC. Since ∠A = 30° in ΔACB, and ∠A = 60° in ΔADB, then ∠A = 30° in ΔADC per angle addition postulate.

Now that we have shown that ΔADB and ΔADC are 30-60-90 triangles, we can use that formula to calculate the side lengths.

CD = 4 cm (given) so AC = 2(4 cm) = 8 cm

Since AC ≅ BC, then BC = 8 cm. Therefore, BD = 4 + 8 = 12 by segment addition postulate.

Lastly, look at ΔBHD. Since ∠B = 30° and ∠H = 90°, then ∠D = 60°. So, ΔBHD is also a 30-60-90 triangle.

BD = 12 cm, so HD = $\frac{12}{2}cm$ = 6 cm

Answer: 6 cm

8 0

3 years ago

Read 2 more answers

Solve for all solutions (2x+3)2=81

Scrat [10]

X=12
Hope this helps

3 0

2 years ago

Read 2 more answers

(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti

Rashid [163]

Answer:

The solution

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}$

Step-by-step explanation:

Explanation:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

Step(i):-

Given differential problem

y′+3 y=9 t

Take the Laplace transform of both sides of the differential equation

L( y′+3 y) = L(9 t)

Using Formula Transform of derivatives

L(y¹(t)) = s y⁻(s)-y(0)

By using Laplace transform formula

$L(t) = \frac{1}{S^{2} }$

Step(ii):-

Given

L( y′(t)) + 3 L (y(t)) = 9 L( t)

$s y^{-} (s) - y(0) + 3y^{-}(s) = \frac{9}{s^{2} }$

$s y^{-} (s) - 7 + 3y^{-}(s) = \frac{9}{s^{2} }$

Taking common y⁻(s) and simplification, we get

$( s + 3)y^{-}(s) = \frac{9}{s^{2} }+7$

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

Step(iii):-

By using partial fractions , we get

$\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}$

$\frac{9}{s^{2} (s+3} = \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}$

On simplification we get

9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Put s =0 in equation(i)

9 = B(0+3)

B = 9/3 = 3

Put s = -3 in equation(i)

9 = C(-3)²

C = 1

Given Equation 9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

9 = A s²+3 A s +B(s)+3 B +C(s²)

0 = A + C

put C=1 , becomes A = -1

$\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}$

Step(iv):-

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

$y^{-}(s) =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}$

Applying inverse Laplace transform on both sides

$L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})$

By using inverse Laplace transform

$L^{-1} (\frac{1}{s} ) =1$

$L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}$

$L^{-1} (\frac{1}{s+a} ) =e^{-at}$

Final answer:-

Now the solution , we get

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}$

5 0

3 years ago

Does 2^8 equal (3sqrt{16})^6

ale4655 [162]

Answer:

Does 2⁸ = (3√16)⁶? No.

Step-by-step explanation:

2⁸ = (3√16)⁶

Evaluate 2⁸. This can simply be written out as:

2 * 2 * 2 * 2 * 2 * 2 * 2 * 2

4 * 2 * 2 * 2 * 2 * 2 * 2

8 * 2 * 2 * 2 * 2 * 2

16 * 2 * 2 * 2 * 2

32 * 2 * 2 * 2

64 * 2 * 2

128 * 2

256.

256 = (3√16)⁶

Evaluate √16.

256 = (3*4)⁶

Multiply 3 and 4.

256 = 12⁶

Evaluate 12⁶. This can be rewritten as:

12 * 12 * 12 * 12 * 12 * 12

144 * 12 * 12 * 12 * 12

1728 * 12 * 12 * 12

20736 * 12 * 12

248832 * 12

2985984.

Since 256 ≠ 2985984, the answer to this question is no.

4 0

2 years ago

Simplify 6(x + 2) + 7.

Elina [12.6K]

Answer:

6x+19

Step-by-step explanation:

First thing first, we need to multiply the 6 into the equation. We do this by multiplying the x and 2 by 6.

6(x+2)+7

Multiply

6x+12+7

Now we need to combine like terms by adding the 12 and 7

6x+19

There it is simplified.

8 0

2 years ago