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2 years ago

M∆abc=17x+8,m∆abk=42°, and m∆kbc=12x-4. Find m∆abc

1 answer:

8 0

So, we have as a given that:
measure angle ABK + measure angle KBC = measure angle ABC

Substitute with the givens and solve for x first.
42 + 12x - 4 = 17x + 8
42 - 4 - 8 = 17x - 12x
30 = 5x
x = 30/5 = 6

Therefore:
measure angle ABC = 17x + 8 = 17(6) + 8 = 110 degrees

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replace each star with a digit to make the problem true.Is there only one answer to each problem? ****-***=2

Semenov [28]

Answer: We have two solutions:

1000 - 998 = 2

1001 - 999 = 2

Step-by-step explanation:

So we have the problem:

****-*** = 2

where each star is a different digit, so in this case, we have a 4 digit number minus a 3 digit number, and the difference is 2.

we know that if we have a number like 99*, we can add a number between 1 and 9 and we will have a 4-digit as a result:

So we could write this as:

1000 - 998 = 2

now, if we add one to each number, the difference will be the same, and the number of digits in each number will remain equal:

1000 - 998 + 1 - 1 = 2

(1000 + 1) - (998 + 1) = 2

1001 - 999 = 2

now, there is a trivial case where we can find other solutions where the digits can be zero, like:

0004 - 0002 = 2

But this is trivial, so we can ignore this case.

Then we have two different solutions.

7 0

2 years ago

What is the simplest form of RootIndex 3 StartRoot 27 a cubed b Superscript 7 Baseline EndRoot?

Vinvika [58]

Answer: $3ab\sqrt[3]{b^4}$

Step-by-step explanation:

Given the following expression:

$\sqrt[3]{27a^3b^7}$

You need to apply the Product of powers property, which states that:

$(a^m)(a^n)=a^{(m+n)$

Then, you can rewrite the expression as following:

$=\sqrt[3]{27a^3b^4b^3}$

The next step is to descompose 27 into its prime factors:

$27=3*3*3=3^3$

Now you must substitute $3^3$ inside the given root. Then:

$=\sqrt[3]{3^3a^3b^4b^3}$

You need to remember that, according to Radicals properties:

$\sqrt[n]{a^n}=a^{\frac{n}{n}}=a^1=a$

Therefore, the final step is to apply this property in order to finally get the expression is its simplest form. This is:

$=3^{\frac{3}{3}}a^{\frac{3}{3}}b^{\frac{4}{3}}b^{\frac{3}{3}}=3ab^{\frac{4}{3}}b=3ab\sqrt[3]{b^4}$

3 0

2 years ago

Read 2 more answers

Suppose we want to choose 2 colors without replacement from the 3 colors

Rzqust [24]

Question:

Suppose we want to choose 3 colors, without replacement, from the red, blue, green, and purple.

(a)How many ways can this be done, if the order of the choices is taken into consideration?

(b)How many ways can this be done, if the order of the choices is not taken into consideration?

Answer:

(a) 6 choices

(b) 3 Choices

Step-by-step explanation:

Given:

3 Colors

Order of choice : Red, Blue, Green

(a) Order of the choices taken into consideration :

When we are picking the first colour there are 3 color choice.

When we are picking second colour there are 2 choice and so on.

Therefore, total choices are = 3*2 = 6 choices

(b) Order of the choices not taken into consideration :

When to take this option we choose as $3 \times 2$ and the remove the duplicates

Therefore, choices are = $\frac{(3 \times 2)}{(1 \times 2)}$ = $\frac{6}{2}$ = 3 Choices

7 0

2 years ago

Tony’s class needs more than $500 for the school dance. So far, they have raised $200. They plan to have a car wash, charging $8

anzhelika [568]

<h2><em><u>PLZ MARK BRAINLIEST</u></em></h2><h2><em><u></u></em></h2>

No, he is not correct for if they washed 37 cars they would have made a total of 296 dollars washing cars. Because this is an inequality, he needs to get a value equal too or greater than 300 dollars on car washes. Also, because he already has 200 dollars collected, you could write the inequality as "8x is greater than or equal to 300".

3 0

2 years ago

Read 2 more answers

Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u

soldier1979 [14.2K]

$x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }$

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

$p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}$

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

$\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}$

$\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}$

$\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}$

Therefore:

$\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]$

Recall that at t=25, $p(25)=\dfrac { 5800 } {13} \approx 446.15$

Therefore:

$\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009$

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

8 0

2 years ago