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diamong [38]
3 years ago
8

Solve this please

\leqslant x + 5" alt="2x + 3 \leqslant x + 5" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Nastasia [14]3 years ago
6 0

Answer:

x ≤ 2

Step-by-step explanation:

Isolate the variable, x. Note the ≤ sign, treat it like an equal sign, what you do to one side, you do to the other.

First, subtract 3 and x from both sides:

2x (-x) + 3 (-3) ≤ x (-x) + 5 (-3)

2x - x ≤ 5 - 3

Simplify:

2x - x ≤ 5 - 3

x ≤ 2

x ≤ 2 is your answer.

~

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Write the explicit form of the arithmetic sequence shown below:<br> -4, 3, 10, ...
Vinil7 [7]

Answer:

17

Step-by-step explanation:

_4 to +3 = 7

3 to 10 =7

thus the sequence is 7

the answer is 10+7=17

7 0
2 years ago
Graph the Quadratic Function given.<br> y = -x^2 - 4x - 3
Dovator [93]

Answer:

Concave down

Y-intercept: (0,-3)

X-intercepts: (-1,0) (-3,0)

Vertex: (-2,1)

6 0
2 years ago
Can someone help me please
IgorLugansk [536]

Answer:

B

Step-by-step explanation:

The slope is 4.

Let's find the slope between your two points.

(1,−2); (3,6)

(x1,y1) = (1,−2)

(x2,y2) = (3,6)

Use the slope formula:

m = y2−y1 / x2−x1

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= 8 / 2

= 4

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7 0
2 years ago
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Only 6 I need need serious help
k0ka [10]

Answer:

100,000+60,000+60

Step-by-step explanation:

7 0
3 years ago
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To watch the video
Anettt [7]

Answer:

The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.

Step-by-step explanation:

Let us take the point of projection of the ball as origin of the coordinate system, the upward direction as positive and down direction as negative.

Initial velocity u with which the ball is projected upwards = + 120 ft/s

Uniform acceleration a acting on the ball is to acceleration due to gravity = - 32 ft/s²

The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.

Using the formula:

v² - u² = 2 a h,

where

u = initial velocity of the ball = +120 ft/s

v = final velocity of the ball at the highest point = 0 ft/s

a = uniform acceleration acting on the ball = -32 ft/s²

h = height attained

Substituting the values we get;

0² - 120² = 2 × (- 32) h

=> h = 120²/2 × 32 = 225 feet

The height of the ball from the ground at its highest point = 225 feet + 12 feet = 237 feet.

7 0
2 years ago
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