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sveticcg [70]
3 years ago
8

2x plus 3 plus x2; for x = 4

Mathematics
2 answers:
Ainat [17]3 years ago
5 0

Answer:

27

Step-by-step explanation:

Not sure if you mean x to the power of 2, so I'll assume that.

Replacing the variable for 4 should look like:

2(4) + 3 + 4^2

8 + 3 + 16 (4 to the power of 2 is equivalent to 4 x 4 which is 16)

Now add them all, 27.

Alborosie3 years ago
4 0

Answer:

2x+3+x2 if x=4 is 27

Step-by-step explanation:

2*x or 2*4=8

8+3=11;

and x or 4 times itself is 16

so we add it all together and get 27.

Have a great day!

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Find the measures of the angles and arcs identified below​
harkovskaia [24]

Answer:

∠a = 42.5°

∠b = 87°

∠c = 174°

∠d = 90°

∠f = 131°

Step-by-step explanation:

∠a = 85°/2 = 42.5°

∠b = 360° -110° - 76° = 174°/2 = 87°

∠c = 360° -110° - 76° = 174°

∠d = 90°

∠f = 360° - 101° - (64° x 2) = 131°

4 0
3 years ago
Complete the square to determine the maximum or minimum value of the function defined by the expression. x2 + 8x + 6
irga5000 [103]

Answer:

Minimum at (-4, -10)

Step-by-step explanation:

x² + 8x + 6

The coefficient of x² is positive, so the parabola opens upward, and the vertex is a minimum.

Subtract the constant from each side

x² + 8x = -6

Square half the coefficient of x

(8/2)² = 4² = 16

Add it to each side of the equation

x² + 8x + 16 = 10

Write the left-hand side as the square of a binomial

(x + 4)² = 10

Subtract 10 from each side of the equation

(x+ 4)² -10 = 0

This is the vertex form of the parabola:

(x - h)² + k = 0,

where (h, k) is the vertex.

h = -4 and k = -10, so the vertex is at (-4, -10).

The Figure below shows your parabola with a minimum at (-4, -10).

4 0
3 years ago
Can someone please help me with this?!
blagie [28]
The answer is 48 because 8 times 9
8 0
3 years ago
use pascal’s triangle or the binomial theorem to find the coefficient on the x term in the expansion of (x+3)^5
Shkiper50 [21]

Answer:

The answer is a maybe hope this helps

Step-by-step explanation:

8 0
3 years ago
In a parallelogram ABCD point M is the midpoint of diagonal AC what is true about diagonal BD
Shalnov [3]
M is also the midpoint of the diagonal BD.

In a parallelogram, the diagonals are perpendicular bisectors of each other. That means that midpoint of one of the diagonals is also the midpoint of the other diagonal.
8 0
3 years ago
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