Ask question

Ask question

All categories

3 years ago

13

Tamara is making a large omelet and needs 1/ 2 cup of chopped tomatoes. She only has a 1 /8 -cup measuring cup. How many times s

hould Tamara fill this measuring cup for the tomatoes?

2 answers:

juin [17]3 years ago

6 0

Answer:

4 times

Step-by-step explanation:

1/2 is the same thing as saying 4/8. 1 is half of 2 just as 4 is half of 8 :)

son4ous [18]3 years ago

3 0

Answer:

I agree with aerodx14

Step-by-step explanation:

You might be interested in

The first term of an Arithmetic progression is 3and the last term is 9, find the number of terms in the progression if the commo

makvit [3.9K]

Answer:

n = 5 terms

Step-by-step explanation:

An = a1 + (n-1)d

9 = 3 + (n-1) 3/2

9 = 3 + 3/2n - 3/2

9 = 3/2 + 3/2n

9 - 3/2 = 3/2n

(15/2) / (3/2) = n

30/6 = n

7 0

2 years ago

Please help quarter ends tomorrow and I don’t know what to do I’m failing all my classes

Nutka1998 [239]

Step-by-step explanation:

x = -10, y = 10

y= -4x+4

m1×m2 = -1

-4 × m2 = -1

m2 = 1/4

y-y1 = m(x-x1)

y-10 = 1/4(x+10)

y-10 = 1/4x +10/4

(×4)

4y-40=x+10

4y=x+50

x-4y+50=0

3 0

2 years ago

Is it possible to draw a triangle with side lengths 4cm 7cm And 9cm

Lana71 [14]

Answer:

it is

possible

Step-by-step explanation:

<u>sum of the two shortest sides must be greater than the third side.</u>

so

4+7 < 9

11<9

condition satisfies ,

hence proved , we can draw

7 0

3 years ago

The lines shown below are perpendicular. If the green line has a slope of 2/5 what is the slope of the red line

Ksenya-84 [330]

In this item, we are not given with the figure but knowing that these lines ought to be perpendicular then we will be able to derive the relationship between the slopes of the line.

The slopes of the perpendicular line are the negative reciprocals of one another. If we represent the slopes of the lines as m₁ and m₂, the relationship can be written in the form,
(m₁)(m₂) = -1

We are given with one of the slopes. To determine the value of the second slope then,
m₂ = -1/m₁
m₂ = -1/(2/5)
m₂ = -5/2

<em>ANSWER: m₂ = -5/2</em>

6 0

2 years ago

Find the exact value of the expression.<br> tan( sin−1 (2/3)− cos−1(1/7))

Sonja [21]

Answer:

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

Step-by-step explanation:

I'm going to use the following identity to help with the difference inside the tangent function there:

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

Let $a=\sin^{-1}(\frac{2}{3})$ .

With some restriction on $a$ this means:

$\sin(a)=\frac{2}{3}$

We need to find $\tan(a)$ .

$\sin^2(a)+\cos^2(a)=1$ is a Pythagorean Identity I will use to find the cosine value and then I will use that the tangent function is the ratio of sine to cosine.

$(\frac{2}{3})^2+\cos^2(a)=1$

$\frac{4}{9}+\cos^2(a)=1$

Subtract 4/9 on both sides:

$\cos^2(a)=\frac{5}{9}$

Take the square root of both sides:

$\cos(a)=\pm \sqrt{\frac{5}{9}}$

$\cos(a)=\pm \frac{\sqrt{5}}{3}$

The cosine value is positive because $a$ is a number between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ because that is the restriction on sine inverse.

So we have $\cos(a)=\frac{\sqrt{5}}{3}$ .

This means that $\tan(a)=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}$ .

Multiplying numerator and denominator by 3 gives us:

$\tan(a)=\frac{2}{\sqrt{5}}$

Rationalizing the denominator by multiplying top and bottom by square root of 5 gives us:

$\tan(a)=\frac{2\sqrt{5}}{5}$

Let's continue on to letting $b=\cos^{-1}(\frac{1}{7})$ .

Let's go ahead and say what the restrictions on $b$ are.

$b$ is a number in between 0 and $\pi$ .

So anyways $b=\cos^{-1}(\frac{1}{7})$ implies $\cos(b)=\frac{1}{7}$ .

Let's use the Pythagorean Identity again I mentioned from before to find the sine value of $b$ .

$\cos^2(b)+\sin^2(b)=1$

$(\frac{1}{7})^2+\sin^2(b)=1$

$\frac{1}{49}+\sin^2(b)=1$

Subtract 1/49 on both sides:

$\sin^2(b)=\frac{48}{49}$

Take the square root of both sides:

$\sin(b)=\pm \sqrt{\frac{48}{49}$

$\sin(b)=\pm \frac{\sqrt{48}}{7}$

$\sin(b)=\pm \frac{\sqrt{16}\sqrt{3}}{7}$

$\sin(b)=\pm \frac{4\sqrt{3}}{7}$

So since $b$ is a number between $0$ and $\pi$ , then sine of this value is positive.

This implies:

$\sin(b)=\frac{4\sqrt{3}}{7}$

So $\tan(b)=\frac{\sin(b)}{\cos(b)}=\frac{\frac{4\sqrt{3}}{7}}{\frac{1}{7}}$ .

Multiplying both top and bottom by 7 gives:

$\frac{4\sqrt{3}}{1}= 4\sqrt{3}$ .

Let's put everything back into the first mentioned identity.

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

$\tan(a-b)=\frac{\frac{2\sqrt{5}}{5}-4\sqrt{3}}{1+\frac{2\sqrt{5}}{5}\cdot 4\sqrt{3}}$

Let's clear the mini-fractions by multiply top and bottom by the least common multiple of the denominators of these mini-fractions. That is, we are multiplying top and bottom by 5:

$\tan(a-b)=\frac{2 \sqrt{5}-20\sqrt{3}}{5+2\sqrt{5}\cdot 4\sqrt{3}}$

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

4 0

3 years ago