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2 years ago

Please help, PhotoMath isn’t helping and I don’t understand

Mathematics

1 answer:

Pie2 years ago

4 0

Answer:

Option B

Step-by-step explanation:

As here,

In all the options processes in 3rd step .

As the bases are same exponents get added. 

And in Option B 3rd step it has been correctly done form of,

${ ({2}^{ \frac{1}{4} } )}^{4} = 4$

Is correct simplification.

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(x+1)(x+8)= can somebody help me on this

goblinko [34]

X^2+8x+x+8 simplifies to x^2+9x+8

8 0

3 years ago

If f(x)=-3x+3, then f-1(x)=

olganol [36]

Answer: f^{-1}(x) = 3-x/3

Step-by-step explanation: Let y = f(x) and rearrange making x the subject, that is...

1. y = - 3x + 3 ( add 3x to both sides )

2. 3x + y = 3 ( subtract y from both sides )

3. 3x = 3 - y ( divide both sides by 3 )

4. x = 3-y/3

Change x back into terms of y

f^{-1}(x) = 3-x/3

6 0

2 years ago

Read 2 more answers

Please help with this I am completely stuck on it

vaieri [72.5K]

Answer:

$f(x)=\sqrt[3]{x-4} , g(x)=6x^{2}\textrm{ or }f(x)=\sqrt[3]{x},g(x)=6x^{2} -4$

Step-by-step explanation:

Given:

The function, $H(x)=\sqrt[3]{6x^{2}-4}$

Solution 1:

Let $f(x)=\sqrt[3]{x}$

If $f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}$ , then,

$\sqrt[3]{g(x)} =\sqrt[3]{6x^{2}-4}\\g(x)=6x^{2}-4$

Solution 2:

Let $f(x)=\sqrt[3]{x-4}$ . Then,

$f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}\\\sqrt[3]{g(x)-4}=\sqrt[3]{6x^{2}-4} \\g(x)-4=6x^{2}-4\\g(x)=6x^{2}$

Similarly, there can be many solutions.

7 0

3 years ago

Solve for the variable: 2x - 28 = 16

svlad2 [7]

2x = 44
44/2 = x
X= 22

4 0

2 years ago

Read 2 more answers

In MON, J, K, and L are midpoints. If JL = 11, LK = 13, and ON = 20, and JL || MN, LK || MO, and JK || ON, what is the length of

In-s [12.5K]

Answer:

The lengths of MN is 22 units, MO is 26 units and JK is 10 units

Step-by-step explanation:

A line segment joining the mid-points of two sides in a triangle is parallel to the third side and equal to half its length

In Δ MON

∵ J, K, and L are mid-points

∵ JL // MN and LK // MO

∴ L is the mid-point of ON

∴ J is the mid-point of MO

∴ K is the mid-point of MN

∵ J, L are the mid-points of MO and ON

∵ JL is opposite to MN

→ By using the rule above

∴ JL = $\frac{1}{2}$ MN

∵ JL = 11 units

∴ 11 = $\frac{1}{2}$ MN

→ Multiply both sides by 2

∴ 22 = MN

∴ MN = 22 units

∵ K, L are the mid-points of MN and ON

∵ KL is opposite to MO

→ By using the rule above

∴ KL = $\frac{1}{2}$ MO

∵ KL = 13 units

∴ 13 = $\frac{1}{2}$ MO

→ Multiply both sides by 2

∴ 26 = MO

∴ MO = 26 units

∵ J, K are the mid-points of MO and MN

∵ JK is opposite to ON

→ By using the rule above

∴ JK = $\frac{1}{2}$ ON

∵ ON =20 units

∴ JK = $\frac{1}{2}$ (20)

∴ JK = 10 units

∴ The lengths of MN are 22 units, MO is 26 units and JK is 10 units

3 0

3 years ago