2. The dishwasher at a restaurant is loaded with the same

Find the quotient. 5 1/4 ÷ 1 3/4. *

weeeeeb [17]

Answer:

3

Step-by-step explanation:

私は怠惰なのでグーグルで調べました

5 0

3 years ago

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What is the measure of Angle

postnew [5]

Answer:

90°

Step-by-step explanation:

90°

8 0

3 years ago

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98 points! plz hurry!

LenaWriter [7]

Answer:

B. (-0.73, 0), (2.73, 0)

Step-by-step explanation:

vertex form of a parabola

y = a(x-h)^2 +k

y = a(x-1)^2 -9

substitute the point in to find a

-6 = a (0-1) ^2 -9

-6 = a *1 -9

add 9 to each side

-6+9 = a -9+9

3 =a

y = 3(x-1)^2 -9

FOIL

y = 3(x-1)(x-1) -9

y = 3(x^2-2x+1) -9

distribute

y = 3x^2-6x+3-9

combine like terms

y = 3x^2 -6x -6

factor out 3

y= 3(x^2 - 2x -2)

set = 0

0 = 3(x^2 - 2x -2)

x^2 - 2x -2 =0

using the quadratic formula

-b±sqrt(b^2 -4ac)

----------------------

2a

-(-2) ± sqrt(2^2 - 4(1)(-2))

-------------------------------

2(1)

2 ± sqrt(4+8)

------------------

2

2±sqrt(12)

---------------

2

2±2sqrt(3)

----------------

2

1±sqrt(3)

roots: 2.73, and -.73

7 0

3 years ago

Prove that:

Lorico [155]

A.)

$\csc^2(x) \tan^2 (x)- 1 = \tan^2(x)$

Use the identities $\csc x = 1 / \sin x$ and $\tan x = \sin x / \cos x$ on the left-hand side

$\begin{aligned}
\text{LHS} &= \csc^2(x) \tan^2 (x)- 1 \\
&= \frac{1}{\sin^2 (x)} \cdot \frac{\sin^2 (x)}{\cos^2 (x)} - 1 \\
&= \frac{1}{\cos^2 (x)} - 1
\end{aligned}$

Make 1 have a common denominator to allow for fraction subtraction
Multiply the numerator and denominator of 1 by cos^2 x

$\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - 1 \cdot \tfrac{\cos^2 (x)}{\cos^2 (x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{\cos^2 (x)}{\cos^2 (x)} \\
&= \frac{1 - \cos^2 x}{\cos^2 (x)}
\end{aligned}$

Use Pythagorean identity for the numerator.

If $\sin^2 (x) + \cos^2(x) = 1$ then subtracting both sides by $\cos^2 (x)$ yields $\sin^2(x) = 1 - \cos^2(x)$ . We can substitute that into the numerator

$\begin{aligned} \text{LHS} &= \frac{1 - \cos^2 (x)}{\cos^2 (x)} \\
&= \frac{\sin^2 (x)}{\cos^2 (x)} \\
&= \tan^2 (x) && \text{Since } \tan x = \tfrac{\sin x }{\cos x} \\
&= \text{RHS}
\end{aligned}$

======

b.)

$\dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} = 1$

For the left-hand side:
By definition, $\sec(x) = 1/\cos(x)$ and $\tan (x) = 1/\cot (x)$

$\begin{aligned}
\text{LHS} &= \dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} \\
&= \dfrac{ \frac{1}{\cos(x)} }{\cos(x)} - \dfrac{\frac{1}{\cot(x)}}{\cot(x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{1}{\cot^2(x)} 
\end{aligned}$

Since $\cot (x) = \cos (x) / \sin (x)$

$\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - \frac{1}{\frac{\cos^2(x)}{\sin^2(x)} } \\ &= \frac{1}{\cos^2 (x)} -\frac{\sin^2(x)}{\cos^2(x)} \\ &= \frac{1 - \sin^2(x)}{\cos^2 (x)} \end{aligned}$

Using Pythagorean identity, $\cos^2(x) = 1 - \sin^2(x)$ so

$\begin{aligned} \text{LHS} &= \frac{\cos^2(x)}{\cos^2 (x)} \\
&= 1 \\
&= \text{RHS}
\end{aligned}$

6 0

3 years ago

Solve 2x2 − 6x = −8.

White raven [17]

2x²-6x+8=0
x²-3x+4=0
D=b²-4ac=9-16=-7
x=(3+/-i√7)/2
correct answer is the last one

3 0

4 years ago

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