Solve 821×54 using an area model

$1.66 x 100 = ? I can't seem to find the correct answer and I must show my work

crimeas [40]

Answer:

166

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

PLS HELP!!!!What are the sine, cosine and tangent ratios of<br>angle W in the triangle below

Ksju [112]

Step-by-step explanation:

Remember SOH-CAH-TOA:

Sine = Opposite / Hypotenuse

Cosine = Adjacent / Hypotenuse

Tangent = Opposite / Adjacent

The side adjacent to W is 4. The side opposite of W is 3. The hypotenuse is 5.

Therefore:

Sine = 3 / 5

Cosine = 4 / 5

Tangent = 3 / 4

7 0

3 years ago

Hayley begins solving this problem by combining like terms. 3x + 5x = 10 Which problem begins the same way? A) 5x = 20 B) 3(x +

joja [24]

Solution: The given equation is →3 x+ 5 x=10

As you can see in the above equation on left side of equation the two terms are variables and on right side of equation the term is a constant term.Also you can see there is an operation called addition between the two variables on left side of equation.

In option (A), there is only a single term on right as well as left,so this can't be true.

In option (B), there is a constant and variable on left side of equation.So this is also untrue.

In option (C), variable being divided by numerator, so this is also untrue.

Option (D)3 x - 2 x = 10, is surely correct , because there are two variables on left side of equation and one constant on right side of equation, and there is an operation called subtraction between them .

6 0

3 years ago

Point e is located at -5,2 point m is the reflection of point e what is the distance between e and m

nataly862011 [7]

Answer:

the distance is 10

Step-by-step explanation:

-5,2 to 0,0 is 5

and the refection of -5,2 is 5,2

5,2 to 0,0. is also 5

5+5=10

4 0

2 years ago

16) Please help with question. WILL MARK BRAINLIEST + 10 POINTS.

Katyanochek1 [597]

We will use the sine and cosine of the sum of two angles, the sine and consine of $\frac{\pi}{2}$ , and the relation of the tangent with the sine and cosine:

$\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta$

$\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0$

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$

If you use those identities, for $\alpha=x,\ \beta=\dfrac{\pi}{2}$ , you get:

$\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x$

$\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$

Hence:

$\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x$

3 0

3 years ago