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Sladkaya [172]
3 years ago
10

Plz help very important

Mathematics
1 answer:
allsm [11]3 years ago
7 0
I hate questions like this. It's like wishing for more wishes.
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Determine whether the triangles are similar by aa , ss , or sas ??
max2010maxim [7]

Answer:

1. AA

2. SSS

3. I´m not sure about this one, I´m sorry

4. Not similar

Step-by-step explanation:

This photo can help you identify!! I hope this helps

4 0
3 years ago
$14,038 is invested, part at 14% and the rest at 12%. If the interest earned from the amount invested at 14% exceeds the interes
Advocard [28]

Answer:

492.99

Step-by-step explanation: x = amount invested at 12%

y = amount invested at 9%

---

x + y = 20287

0.12x = 0.09y + 492.99

---

put the system of linear equations into standard form

---

x + y = 20287

0.12x - 0.09y = 492.99

5 0
2 years ago
Find y’ for y=7sec^3x
laiz [17]
Taking the derivative of 7 times secant of x^3:
We take out 7 as a constant focus on secant (x^3)
To take the derivative, we use the chain rule, taking the derivative of the inside, bringing it out, and then the derivative of the original function. For example:
The derivative of x^3 is 3x^2, and the derivative of secant is tan(x) and sec(x).
Knowing this: secant (x^3) becomes tan(x^3) * sec(x^3) * 3x^2. We transform tan(x^3) into sin(x^3)/cos(x^3) since tan(x) = sin(x)/cos(x). Then secant(x^3) becomes 1/cos(x^3) since the secant is the reciprocal of the cosine.

We then multiply everything together to simplify:

sin(x^3) * 3x^2/ cos(x^3) * cos(x^3) becomes

3x^2 * sin(x^3)/(cos(x^3))^2

and multiplying the constant 7 from the beginning:

7 * 3x^2 = 21x^2, so...

our derivative is 21x^2 * sin(x^3)/(cos(x^3))^2


6 0
3 years ago
If you can answer any of these please do and put the question number by the answer
Ronch [10]
19. 5x - 5 = 3x -9
2x = -4
x = -2

19. Answer is A
8 0
3 years ago
28 students in the class 75% pass how many students pass
SashulF [63]

Answer:

21 students pass

Step-by-step explanation:

Firstly, you can set up the problem into an equation where the variable X would equal the number of students passing. You put X over the total number of students in the class, turning it into a fraction, then set it equal to the fraction  \frac{75}{100} (which is 75% represented as a fraction).

\frac{X}{28} = \frac{75}{100}

The fraction \frac{75}{100} can be simplified, because 75 and 100 are both multiples of 25, so after canceling out the 25s you would be left with \frac{3}{4}.

\frac{X}{28} = \frac{3}{4}

Next, you use the process of cross multiplication which is essentially just multiplying the denominators of both fractions (which would be 28 and 4 in this case) to each side of the equation.

\frac{X}{28} * 28 * 4 = \frac{3}{4} * 4 * 28

The denominators cancel out leaving you with a simple equation to simplify.

4* X =  3 * 28

4X = 84

Finally, divide both sides by four in order to isolate the variable.

\frac{4X}{4} = \frac{84}{4}

X = 21.

5 0
3 years ago
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