Pleaseee helpp me i got one down!!! (zoom in to see more clear)

Hello there!

Before we answer the question, we must solve the equation first.

4x + 5 = 10

4x = 10 - 5

4x = 5

x = 5/4

Therefore,

The equation has one solution!

5 0

2 years ago

This graph is an example of

masha68 [24]

Answer:

option A is correct. this graph is a linear function

Step-by-step explanation:

Linear functions are those whose graph is a straight line. A linear function has one independent variable and one dependent variable. The independent variable is x and the dependent variable is y. a is the constant term or the y intercept

7 0

2 years ago

Read 2 more answers

Integral of 1/(x^2sqrt(4x+1))?

cestrela7 [59]

Integral of 1/(x^2sqrt(4x+1)) = 4tanh^-1(sqrt(4x + 1)) - sqrt(4x + 1) / x + c

3 0

2 years ago

Felix earns $1,400 a month. His cell phone represents 5% of his income. How much is his cell phone cost each month?

solong [7]

I believe is 70 multiply 1400 times 5% and it will give you 70

4 0

3 years ago

Read 2 more answers

Plz help I have no idea how to answer it

saveliy_v [14]

5. Answer: see explanation

Step-by-step explanation:

If the roots are m and 3m, then x = m and x = 3m

⇒ x - m = 0 and x - 3m = 0

⇒ (x - m)(x - 3m) = 0

⇒ x² - 4mx + 3m² = 0

Since x² + px + q = 0 then p = -4m and q = 3m²

3p² = 3(-4m)² 16q = 16(3m²)

= 3(16m²) = 48m²

= 48m²

3p² = 48m² = 16q ⇒ 3p² = 16q

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6. Answer: 8 or 18

Step-by-step explanation:

The Area of the entire rectangle (A = L × w) is 12 × 10 = 120

The Area of the shaded region is 72, so the Area of the non-shaded region is 120 - 72 = 48.

There are two non-shaded triangles.

Bigger non-shaded Δ: L = 12-x, w = 10 ⇒ $A = \dfrac{10(12-x)}{2}$
Smaller non-shaded Δ: L = x, w = x ⇒ $A=\dfrac{x(x)}{2}$

Combine the Areas of both triangles and set it equal to the Area of the non-shaded region:

$\dfrac{x(x)}{2}+\dfrac{10(12-x)}{2}=48\\\\\\\dfrac{x^2}{2}+\dfrac{120-10x}{2}=48\\\\\\\dfrac{x^2-10x+120}{2}=48\\\\\\x^2-10x+120=96\\\\x^2-10x+24=0\\\\(x-4)(x-6)=0\\\\x - 4 = 0\quad or\quad x-6=0\\x=4\qquad or\qquad x=6\\\\\text{Based on the image provided, it appears that the x-value will be the}\\\text{smallest of the two possible solutions (4), but both solutions are valid.}$

Area of ΔBEF:

$\text{when x = 4, A =}\ \dfrac{4(4)}{2}=\dfrac{16}{2}=8\\\\\\\text{when x = 6, A =}\ \dfrac{6(6)}{2}=\dfrac{36}{2}=18$

5 0

2 years ago