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gayaneshka [121]
3 years ago
7

If f(x) = 2x2 − 10, find f(5). 90 10 7.5 40

Mathematics
2 answers:
777dan777 [17]3 years ago
8 0
If you mean 2x^2 (squared) - 10, then all you have to do is replace x with 5.

f(5) = 2(5)^2 - 10
= 2 (25) - 10
= 50 - 10
f(5) = 40
MatroZZZ [7]3 years ago
5 0
F(x)= 2x² -10
f(5)= 2(5)²-10

f(5)= 40
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Rainbow [258]

We know that

x = \sqrt{\frac{4}{16}}

So:

x = \frac{2}{4}

6 0
3 years ago
If 111 is divided into pieces that are each \dfrac19
PIT_PIT [208]

The results of dividing 1 into pieces that are each 1/9 of a whole is 9

<h3>Division of fraction</h3>

1 ÷ 1/9

  • multiply by the reciprocal of 1/9 which is 9/1

1 ÷ 1/9

= 1 × 9/1

= (1 × 9) / (1 × 1)

= 9/1

= 9

Complete question:

If 1 is divided into pieces that are each 1/9 of a whole, how many pieces are there

Learn more about fraction:

brainly.com/question/1622425

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8 0
1 year ago
Find the projection of u onto v<br> URGENT HELP
padilas [110]

Given:

Two vectors are:

u=\left

v=\left

To find:

The projection of u onto v.

Solution:

Magnitude of a vector v=\left is:

|v|=\sqrt{a^2+b^2}

Dot product of two vector v_1=\left and v_2=\left is:

v_1\cdot v_2=a_1a_2+b_1b_2

Formula for projection of u onto v is:

Proj_vu=\dfrac{u\cdot v}{|v|^2}v

Proj_vu=\dfrac{\left \cdot \left }{(\sqrt{2^2+17^2)^2}}\left

Proj_vu=\dfrac{0\cdot 2+6\cdot 17}{4+289}\left

Proj_vu=\dfrac{0+102}{293}\left

On further simplification, we get

Proj_vu=\dfrac{102}{293}\left

Proj_vu=\left

Proj_vu=\left

Therefore, the projection of u onto v is Proj_vu=\left.

3 0
3 years ago
Find the area of a triangle whose vertices are (0,0), (4,2), (-1,2)​
Yakvenalex [24]

Answer:

Therefore area of a triangle whose vertices are (0,0), (4,2), (-1,2) is

5 units².

Step-by-step explanation:

Given:

Let the vertices be,

point A( x₁ , y₁) ≡ ( 0 , 0)

point B( x₂ , y₂) ≡ (4 , 2)

point C(x₃ , y₃ ) ≡ (-1 , 2)

To Find:

Area of Triangle = ?

Solution:

If the Vertices A( x₁ , y₁), B( x₂ , y₂) and C(x₃ , y₃ ) then the Area of Triangle is given by

\textrm{Area of Triangle}=\dfrac{1}{2}(x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}))

Substituting the values we get

\textrm{Area of Triangle}=\dfrac{1}{2}(0(2-2)+4(2-0)+(-1)(0-2))

\textrm{Area of Triangle}=\dfrac{1}{2}(0+8+2))=\dfrac{10}{2}=5\ units^{2}

Therefore area of a triangle whose vertices are (0,0), (4,2), (-1,2) is

5 units².

7 0
3 years ago
Answer any of these plz. AND HURRY!!!
faust18 [17]
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