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3 years ago

Please Help! Ill give brainliest no fakes PLEASE!!

Mathematics

2 answers:

valina [46]3 years ago

7 0

second one: 6:4

last one: 15:10

9966 [12]3 years ago

3 0

Answer:

4 and then 15

Step-by-step explanation:

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Picture attached please help data handling <br>i need help for question (b), (c) and d

Elis [28]

Answer:

86% for b

Step-by-step explanation:

3 0

3 years ago

Translate the expression with rational exponents. Simplify if necessary

Misha Larkins [42]

Answer: 27

Step-by-step explanation:

The expression given is inside a cube root. This is a simplified version of a fractional exponent, 1/3.

We can take the expression outside of the radical to get this:

3 ^ 9 ^ (1/3)

When an exponent is raised to another exponent, multiply the exponents and keep the base the same:

3 ^ (9 * 1/3)

3 ^ 3

The expression is equal to 27.

5 0

3 years ago

Consider the system of equations. y = 3x + 2 y = − 2 3 x − 4 Explain why these particular equations can be graphed immedia

inessss [21]

Answer:

These equations are in slope-intercept form. I can use the y-intercept and slope to graph both lines. I plot the y-intercept and use rise over run to locate another point on the line. Then, I can draw a line through the two points.

Step-by-step explanation:

3 0

3 years ago

I need help on all of these plz

shusha [124]

Can't read the top questions mate !

8 0

3 years ago

Verify cot x sec^4x=cotx +2tanx +tan^3x

Tanzania [10]

Answer:

See explanation

Step-by-step explanation:

We want to verify that:

$\cot(x) \: { \sec}^{4} x = \cot(x) + 2 \tan(x) + { \tan}^{3} x$

Verifying from left, we have

$\cot(x) \: { \sec}^{4} x = \cot(x) \: ( 1 + { \tan}^{2} x )^{2}$

Expand the perfect square in the right:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: ( 1 + { 2\tan}^{2} x + { \tan}^{4} x)$

We expand to get:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: + \cot(x){ 2\tan}^{2} x +\cot(x) { \tan}^{4} x$

We simplify to get:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: + 2 \frac{ \cos(x) }{\sin(x) ) } \times \frac{{ \sin}^{2} x}{{ \cos}^{2} x} +\frac{ \cos(x) }{\sin(x) ) } \times \frac{{ \sin}^{4} x}{{ \cos}^{4} x}$

Cancel common factors:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: + 2 \frac{{ \sin}x}{{ \cos}x} +\frac{{ \sin}^{3} x}{{ \cos}^{3} x}$

This finally gives:

$\cot(x) \: { \sec}^{4} x = \cot(x) + 2 \tan(x) + { \tan}^{3} x$

3 0

3 years ago