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4 years ago

Please answer this correctly without making mistakes

Mathematics

2 answers:

Lelu [443]4 years ago

8 0

Answer:

so first convert to fraction so

9 3/4 = 39/4

so it was spread among 3

so this is division so you do 39/4 divided by 3

so you keep switch flip

which is 39/4 *1/3

answer is 13/4

trasher [3.6K]4 years ago

7 0

Answer:

Step-by-step explanation:

$9\frac{3}{4}= \frac{(4 \times 9)+3}{4}= \frac{39}{4} \\\\\frac{39}{4} = 3 \:vegetable \: beds\\x \:\:\:= 1 \: vegetable \:bed\\\\3x = \frac{39}{4} \\\\\frac{3x}{3} = \frac{\frac{39}{4} }{3} \\\\x = \frac{13}{4} \\\\x = 3\frac{1}{4}$

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What value of x makes this equation true? 6x - 3 = 5x - 5

malfutka [58]

Answer:

x= - 2

Step-by-step explanation:

3 0

3 years ago

Maria, bill, and change sent a total of 71 text messages during the weekend. change sent 2 times as many messages as bill. maria

VLD [36.1K]

Alright, we're dealing with a few values here, so let's give them some labels to save us some trouble down the road. We'll call the number of messages sent by Maria m, the number sent by Bill b and the number sent by Change (is that a real name?) c. We don't know exactly what each number is, but let's take a look at what information they do give us.

Change sent 2 times as many messages as Bill, or, using our variable for Change and Bill:

$c=2b$

We're also given that Maria sent 7 messages more than Bill, which we can represent with:

$m=b+7$

Notice that m and c are both in terms of b. We can use this for our next step. We're given at the beginning that together, Maria, Bill, and Change sent 71 messages over the weekend. As an equation using all of our variables, this translates to:

$m+b+c=71$

Since m and c are both in terms of b, we can substitute those expressions in and solve for b:

$(b+7)+b+2b=71\\ b+7+b+2b=71\\ 7+4b=71\\ 4b=64\\ b=16$

Now that know that Bill sent 16 texts, we can find the numbers for Change and Maria:

$m=b+7=16+7=23\\
c=2b=2(16)=32$

So, Bill sent 16 texts, Maria sent 23, and Change sent 32.

4 0

3 years ago

Find cot and cos If sec = -3 and sin 0 > 0

Natali5045456 [20]

Answer:

Second answer

Step-by-step explanation:

We are given $\displaystyle \large{\sec \theta = -3}$ and $\displaystyle \large{\sin \theta > 0}$ . What we have to find are $\displaystyle \large{\cot \theta}$ and $\displaystyle \large{\cos \theta}$ .

First, convert $\displaystyle \large{\sec \theta}$ to $\displaystyle \large{\frac{1}{\cos \theta}}$ via trigonometric identity. That gives us a new equation in form of $\displaystyle \large{\cos \theta}$ :

$\displaystyle \large{\frac{1}{\cos \theta} = -3}$

Multiply $\displaystyle \large{\cos \theta}$ both sides to get rid of the denominator.

$\displaystyle \large{\frac{1}{\cos \theta} \cdot \cos \theta = -3 \cos \theta}\\\displaystyle \large{1=-3 \cos \theta}$

Then divide both sides by -3 to get $\displaystyle \large{\cos \theta}$ .

Hence, $\displaystyle \large{\boxed{\cos \theta = - \frac{1}{3}}}$

__________________________________________________________

Next, to find $\displaystyle \large{\cot \theta}$ , convert it to $\displaystyle \large{\frac{1}{\tan \theta}}$ via trigonometric identity. Then we have to convert $\displaystyle \large{\tan \theta}$ to $\displaystyle \large{\frac{\sin \theta}{\cos \theta}}$ via another trigonometric identity. That gives us:

$\displaystyle \large{\frac{1}{\frac{\sin \theta}{\cos \theta}}}\\\displaystyle \large{\frac{\cos \theta}{\sin \theta}$

It seems that we do not know what $\displaystyle \large{\sin \theta}$ is but we can find it by using the identity $\displaystyle \large{\sin \theta = \sqrt{1-\cos ^2 \theta}}$ for $\displaystyle \large{\sin \theta > 0}$ .

From $\displaystyle \large{\cos \theta = -\frac{1}{3}}$ then $\displaystyle \large{\cos ^2 \theta = \frac{1}{9}}$ .

Therefore:

$\displaystyle \large{\sin \theta=\sqrt{1-\frac{1}{9}}}\\\displaystyle \large{\sin \theta = \sqrt{\frac{9}{9}-\frac{1}{9}}}\\\displaystyle \large{\sin \theta = \sqrt{\frac{8}{9}}}$

Then use the surd property to evaluate the square root.

Hence, $\displaystyle \large{\boxed{\sin \theta=\frac{2\sqrt{2}}{3}}}$

Now that we know what $\displaystyle \large{\sin \theta}$ is. We can evaluate $\displaystyle \large{\frac{\cos \theta}{\sin \theta}}$ which is another form or identity of $\displaystyle \large{\cot \theta}$ .

From the boxed values of $\displaystyle \large{\cos \theta}$ and $\displaystyle \large{\sin \theta}$ :-

$\displaystyle \large{\cot \theta = \frac{\cos \theta}{\sin \theta}}\\\displaystyle \large{\cot \theta = \frac{-\frac{1}{3}}{\frac{2\sqrt{2}}{3}}}\\\displaystyle \large{\cot \theta=-\frac{1}{3} \cdot \frac{3}{2\sqrt{2}}}\\\displaystyle \large{\cot \theta=-\frac{1}{2\sqrt{2}}$

Then rationalize the value by multiplying both numerator and denominator with the denominator.

$\displaystyle \large{\cot \theta = -\frac{1 \cdot 2\sqrt{2}}{2\sqrt{2} \cdot 2\sqrt{2}}}\\\displaystyle \large{\cot \theta = -\frac{2\sqrt{2}}{8}}\\\displaystyle \large{\cot \theta = -\frac{\sqrt{2}}{4}}$

Hence, $\displaystyle \large{\boxed{\cot \theta = -\frac{\sqrt{2}}{4}}}$

Therefore, the second choice is the answer.

__________________________________________________________

Summary

Trigonometric Identity

$\displaystyle \large{\sec \theta = \frac{1}{\cos \theta}}\\ \displaystyle \large{\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}}\\ \displaystyle \large{\sin \theta = \sqrt{1-\cos ^2 \theta} \ \ \ (\sin \theta > 0)}\\ \displaystyle \large{\tan \theta = \frac{\sin \theta}{\cos \theta}}$