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3 years ago

What is main difference between the SAT-I and SAT-II?

Mathematics

1 answer:

Wewaii [24]3 years ago

7 0

Sat is the regular sat and sat II is the sat subject tests

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Find the degree measures of the next two positive and the previous two negative angles that are coterminal with the angle 75°.

AlladinOne [14]

9514 1404 393

Answer:

435°, 795°
-285°, -645°

Step-by-step explanation:

Add or subtract multiples of 360° to find coterminal angles.

<u>Positive</u>

75° +360° = 435°

75° +2×360° = 795°

<u>Negative</u>

75° -360° = -285°

75° -2×360° = -645°

5 0

3 years ago

The graph of f(x)=cos(x) is transformed into a new function, g(x) , by stretching it vertically by a factor of 3 and shifting it

tresset_1 [31]

Answer:

g(x) = 3 cos (x+2)

Step-by-step explanation:

To transform functions use the following formula f(x) = a cos b(x -c) + d where:

a stretches the function vertically
b compresses or stretches it horizontally
c shifts the function left or right
d shifts the function up or down

Here a = 3 and c = 2.

So the new function is g(x) = 3 cos (x+2).

5 0

3 years ago

Simplify this: log√24+ log6

Kitty [74]

simplifying $log\sqrt{24}+log6$ we get $log(12\sqrt{6})$

Step-by-step explanation:

We need to simplify: $log\sqrt{24}+log6$

Solving:

Applying the rule:

log a + log b = log(ab)

$log\sqrt{24}+log6\\=log(6\sqrt{24})\\\sqrt{24} = \sqrt{2*2*2*3} =\sqrt{2^2*6} =\sqrt{2^2} \sqrt{6}=2\sqrt{6}\\ Putting\,\,value:\\=log(6*2\sqrt{6})\\=log(12\sqrt{6})$

So, simplifying $log\sqrt{24}+log6$ we get $log(12\sqrt{6})$

Keywords: Simplifying Logarithms

Learn more about Simplifying Logarithms at:

#learnwithBrainly

3 0

3 years ago

Find all values of c such that c/(c - 5) = 4/(c - 4). If you find more than one solution, then list the solutions you find separ

Deffense [45]

Answer:

$\large\boxed{\bold{NO\ REAL\ SOLUTIONS}}\\\boxed{x=4-2i,\ x=4+2i}$

Step-by-step explanation:

$Domain:\\c-4\neq0\ \wedge\ c-5\neq0\Rightarrow c\neq4\ \wedge\ c\neq5\\\\\dfrac{c}{c-5}=\dfrac{4}{c-4}\qquad\text{cross multiply}\\\\c(c-4)=4(c-5)\qquad\text{use the distributive property}\\\\c^2-4c=4c-20\qquad\text{subtract}\ 4c\ \text{from both sides}\\\\c^2-8c=-20\qquad\text{add 20 to both sides}\\\\c^2-8c+20=0\qquad\text{use the quadratic formula}$

$\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x^2-8x+20=0\\\\a=1,\ b=-8,\ c=20\\\\b^2-4ac=(-8)^2-4(1)(20)=64-80=-16$

$\text{In the set of complex numbers:}\\\\i=\sqrt{-1}\\\\\text{therefore}\ \sqrt{b^2-4ac}=\sqrt{-16}=\sqrt{(16)(-1)}=\sqrt{16}\cdot\sqrt{-1}=4i\\\\x=\dfrac{-(-8)\pm4i}{2(1)}=\dfrac{8\pm4i}{2}=\dfrac{8}{2}\pm\dfrac{4i}{2}=4\pm2i$

6 0

3 years ago

Read 2 more answers

How do you know the correct order in which to evaluate algebraic expressions?

lys-0071 [83]

In solving algebraic expressions, there is an order in which to solve it.

1) perform the operation inside the parenthesis first
2) then perform the operation with the exponents
3) apply multiplication and division from left to right of the equation.
4) apply addition and subtraction from left to right of the equation.

Remember this: PEMDAS for easier recall. (parenthesis, exponents, multiplication, division, addition, subtraction)

3 0

3 years ago

Read 2 more answers