Example

Jam is made from sugar and strawberries in the ratio of 3:5 A jar contains 150g of sugar how many grams of strawberries are in t

daser333 [38]

Answer:250

Step-by-step explanation:

150/3 =50

50x5=250

3 0

3 years ago

Read 2 more answers

Write 120 as a percentage simplest form

netineya [11]

Convert the decimal by multiplying by 100.

=12000%

4 0

3 years ago

How many solutions does the system y = -7x + 3 and y + 7x = 10 have? Explain.

Vaselesa [24]

Answer:

i believe its infinitely

Step-by-step explanation:

i say this because the equation is similar on both sides

4 0

3 years ago

The length of a rectangle is five inches less than twice it’s width. If the rectangle has a perimeter of 80 inches, find the len

yawa3891 [41]

Answer:

Width of the rectangle = 15 inches

Length of the rectangle = 25 inches

Step-by-step explanation:

Let the width of the rectangle = k inches

So, the length of the rectangle = (2k -5) inches

Perimeter = 80 in

Perimeter of a rectangle = 2 (Length + Width)

⇒ 80 = 2(k + (2k -5)

or 80 = 6k - 10

⇒ 6k = 80 + 10 = 90

or k = 90 / 6

⇒ k = 15

So the width of the rectangle = k = 15 inches

And the length of the rectangle = 2k -5 = 2 x 25 - 5 = 25 inches

5 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