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3 years ago

Given right triangle ABC with altitude BD drawn to hypotenuse AC. If AB = 8

Mathematics

1 answer:

Delicious77 [7]3 years ago

4 0

Answer:

x = 32

Step-by-step explanation:

∠BCA = ∠DBA (90 - ∠DBC)

∠A = ∠A

ΔABD similar to ΔACB

AC/AB = AB/AD

x / 8 = 8 / 2

x = 32

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For the equation 3x - 5y = -3, what is the value ofy when x is 1?<br> 6/5<br> -6/5<br> 5/6

Law Incorporation [45]

Answer:

y = 6/5

Step-by-step explanation:

3x - 5y = -3

Let x =1

3(1) - 5y = -3

3-5y = -3

Subtract 3 from each side

3-3-5y = -3-3

-5y = -6

Divide each side by -5

-5y/-5 = -6/-5

y = 6/5

8 0

3 years ago

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Can someone help me with this two problems?

ANEK [815]

2x+4y=0

substitute y with 0

2x+4(0)=0

solve the equation

2x+0=0

2x=0

divide by 2 on both sides

x=0

4x+8y=7

substitute y with 0

4x+8(0)=7

solve the equation

4x+0=7

4x=7

divide by 4 on both sides

x=7/4 or x=1 3/4 or x=1.75

3x-7y=-29

2x+2y=6

solve the bottom equation

3x-7y=-29

x=3-y

substitute for x

3(3-y)-7y=-29

solve the equation

y=19/5

now substitute for y

x=3- $\frac{19}{5}$

solve for x

x=-4/5

the possible solution of the system is the ordered pair

(x,y)=( $-\frac{4}{5} ,\frac{19}{5}$ )

3 0

4 years ago

Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |

dsp73

Looks like $f(x,y)=x^2+y^2-x^2y+7$ .

$f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1$

$f_y=2y-x^2=0\implies2y=x^2$

If $x=0$ , then $y=0$ - critical point at (0, 0).
If $y=1$ , then $x=\pm\sqrt2$ - two critical points at $(-\sqrt2,1)$ and $(\sqrt2,1)$

The latter two critical points occur outside of $D$ since $|\pm\sqrt2|>1$ so we ignore those points.

The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}$

The value of its determinant at (0, 0) is $\det H(0,0)=4>0$ , which means a minimum occurs at the point, and we have $f(0,0)=7$ .

Now consider each boundary:

If $x=1$ , then

$f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4$

which has 3 extreme values over the interval $-1\le y\le1$ of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

If $x=-1$ , then

$f(-1,y)=8-y+y^2$

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

If $y=1$ , then

$f(x,1)=8$

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

If $y=-1$ , then

$f(x,-1)=2x^2+8$

which has 3 extreme values, but the previous cases already include them.

Hence $f(x,y)$ has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

3 0

3 years ago

Which of the following words mean subtraction? Check all that apply.

anastassius [24]

Answer is C please give brainliest

5 0

3 years ago

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Lester worked 12 hours last week at the grocery store and earned $93.00. If he continues to earn the same hourly pay, how many a

kykrilka [37]

93 / 12 = 7.75

so Lester earns $7.75 a hour.

if he earns $62 how many hours did he work.

so 62 = 7.75 (h)

62/ 7.75 = 8

so he worked 8 hours.

4 0

3 years ago

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