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3 years ago

7.4 + 8.2 – 3.2 – 2.4 Rewrite subtraction using the additive inverse.

Mathematics

1 answer:

Debora [2.8K]3 years ago

5 0

The answer to 7.4 + 8.2 - 3.2 - 2.4 is 10.

Sadly I do not know how to rewrite subtraction using the additive inverse

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2 years ago

Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list. Enter

Levart [38]

Answer:

Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank

A . tan(arcsin(x/8))

B . cos (arsin (x/8))

C. (1/2)sin (2arcsin (x/8))

D . sin ( arctan (x/8))

E. cos (arctan (x/8))

These are the spaces to fill out :

.. ..........x/64 (sqrt(64-x^2))

.............x/sqrt(64+x^2)

.............sqrt(64-x^2)/8

..............x/sqrt(64-x^2)

..............8/sqrt(64+x^2)

A. ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)

B . cos (arsin (x/8)) ....sqrt(64-x^2)/8

Step-by-step explanation:

To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus

A we have the sides of the triangle given by x, 8 and $\sqrt{8^{2} - x^{2} }$ or $\sqrt{64 - x^{2} }$

thus tan(arcsin(x/8)) = $\frac{x}{\sqrt{64 - x^{2} }}$ =

Therefore ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)

Here we have cos = adjacent/hypotenuse where adjacent side is $\sqrt{64 - x^{2} }$ and hypothenuse = 8 we have $\sqrt{64 - x^{2} }$ /8

B . cos (arsin (x/8)) ....sqrt(64-x^2)/8

4 0

3 years ago

B. Determine the values of a and b so that f is continuous.<br><br> Please help!

wlad13 [49]

Answer:

following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

$\lim_{x\to 2}+f(x) = \lim_{x\to 2}+ (ax^2-bx+3) = 4a-2b+3\\\\ \to 4a-2b+3 = 4\\\\ \therefore \ \ 4a-2b = 1\\\\$

The one-sided limits of F(x) at x = 3 must be equivalent for f(x) to be continuous at x = 3.

$\to \lim_{x\to 3}- f(x) = \lim_{x\to 3}- (ax^2-bx+3) = 9a-3b+3\\\\ \to \lim_{x\to 3}+ f(x) = \lim_{x\to 3}+ (2x-a+b) = 6-a+b\\\\$

So,

$\to 9a-3b+3 = 6-a+b\\\\\therefore\\ \to 10a -4b = 3$

$\to 4a - 2b = 1......(a)\\\to 10a - 4b = 3.......(b)\\$

In equation a multiply the by -2 and then add in the equation b:

$\to -8a + 4b = -2\\\to 10a - 4b = 3\\ \to 2a = 1\\\\ \to a = \frac{1}{2}\\\\ \to \ 4(\frac{1}{2}) - 2b = 1\\\\ \to 2 - 2b = 1\\\\ \to -2b = -1\\\\ \to b = \frac{1}{2}$

So, the value of $a \ and \ b= \frac{1}{2}$

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3 years ago

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There are 2 points of intersection

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