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2 years ago

Which is an example of an expression? O 3,2 – 4. O a = 144 0 10 = 2.5 O 2 < 52

Mathematics

1 answer:

cluponka [151]2 years ago

4 0

Answer:

2 - 4

Step-by-step explanation:

no equal sign

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Triangle ABC is the pre-image and triangle DFE is the image.

anzhelika [568]

Answer:

Scale factor = ½

Step-by-step explanation:

The original image is the preimage = ∆ACB

The new image is the image = ∆DFE

The scale of factor of dilation = image/preimage = DF/AC

DF = 6 cm (given)

AC = 12 cm (given)

Plug in the values into the equation to find the scale factor of dilation:

Scale factor = 6/12

Scale factor = ½

6 0

3 years ago

Find the area of a triangle bounded by the y-axis, the line f(x)=9−4/7x, and the line perpendicular to f(x) that passes through

Setler79 [48]

<u>ANSWER: </u>

The area of the triangle bounded by the y-axis is $\frac{7938}{4225} \sqrt{65} \text { unit }^{2}$

<u>SOLUTION:</u>

Given, $f(x)=9-\frac{-4}{7} x$

Consider f(x) = y. Hence we get

$f(x)=9-\frac{-4}{7} x$ --- eqn 1

$y=9-\frac{4}{7} x$

On rewriting the terms we get

4x + 7y – 63 = 0

As the triangle is bounded by two perpendicular lines, it is an right angle triangle with y-axis as hypotenuse.

Area of right angle triangle = $\frac{1}{ab}$ where a, b are lengths of sides other than hypotenuse.

So, we need find length of f(x) and its perpendicular line.

First let us find perpendicular line equation.

Slope of f(x) = $\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-4}{7}$

So, slope of perpendicular line = $\frac{-1}{\text {slope of } f(x)}=\frac{7}{4}$

Perpendicular line is passing through origin(0,0).So by using point slope formula,

$y-y_{1}=m\left(x-x_{1}\right)$

Where m is the slope and $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$

$y-0=\frac{7}{4}(x-0)$

$y=\frac{7}{4} x$ --- eqn 2

4y = 7x

7x – 4y = 0

now, let us find the vertices of triangle, one of them is origin, second one is point of intersection of y-axis and f(x)

for points on y-axis x will be zero, to get y value, put x =0 int f(x)

0 + 7y – 63 = 0

7y = 63

y = 9

Hence, the point of intersection is (0, 9)

Third vertex is point of intersection of f(x) and its perpendicular line.

So, solve (1) and (2)

$\begin{array}{l}{9-\frac{4}{7} x=\frac{7}{4} x} \\\\ {9 \times 4-\frac{4 \times 4}{7} x=7 x} \\\\ {36 \times 7-16 x=7 \times 7 x} \\\\ {252-16 x=49 x} \\\\ {49 x+16 x=252} \\\\ {65 x=252} \\\\ {x=\frac{252}{65}}\end{array}$

Put x value in (2)

$\begin{array}{l}{y=\frac{7}{4} \times \frac{252}{65}} \\\\ {y=\frac{441}{65}}\end{array}$

So, the point of intersection is $\left(\frac{252}{65}, \frac{441}{65}\right)$

Length of f(x) is distance between $\left(\frac{252}{65}, \frac{441}{65}\right)$ and (0,9)

$\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(9-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+0} \\ &=\frac{252}{65} \end{aligned}$

Now, length of perpendicular of f(x) is distance between $\left(\frac{252}{65}, \frac{441}{65}\right) \text { and }(0,0)$

$\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(0-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+\left(\frac{441}{65}\right)^{2}} \\ &=\frac{\sqrt{(12 \times 21)^{2}+(21 \times 21)^{2}}}{65} \\ &=\frac{63}{65} \sqrt{65} \end{aligned}$

Now, area of right angle triangle = $\frac{1}{2} \times \frac{252}{65} \times \frac{63}{65} \sqrt{65}$

$=\frac{7938}{4225} \sqrt{65} \text { unit }^{2}$

Hence, the area of the triangle is $\frac{7938}{4225} \sqrt{65} \text { unit }^{2}$

8 0

3 years ago

Solve the equation using square roots.

anastassius [24]

Answer:

A : ±2

Step-by-step explanation:

Add 14 and you have ...

... x² = 4

Take the square root and you have ...

... x = ±√4 = ±2

3 0

3 years ago

Let $a$ and $b$ be nonzero real numbers such that

algol [13]

<h3>Answer: -7/2</h3>

=========================================================

Explanation:

Let's expand out (2 - 7i)(a + bi) using the FOIL rule

(2 - 7i)(a + bi) = 2a + 2bi - 7ai - 7bi^2

(2 - 7i)(a + bi) = 2a + 2bi - 7ai - 7b(-1)

(2 - 7i)(a + bi) = 2a + 2bi - 7ai + 7b

(2 - 7i)(a + bi) = (2a+7b) + (2bi-7ai)

(2 - 7i)(a + bi) = (2a+7b) + (2b-7a)i

We're told the result is purely imaginary. What this means is that the real part (2a+7b) is zero, while the imaginary part (2b-7a) is nonzero. If both are zero, then we have 0+0i = 0 which is purely real.

For example, the complex numbers 0-7i and 0+2i are purely imaginary.

Let's use the fact that 2a+7b must be zero to do the following steps:

2a+7b = 0

2a = -7b

a = -7b/2

a/b = -7/2 which is the final answer

We must check to see if 2b-7a is nonzero

2b - 7a = 2b - 7(-7b/2)

2b - 7a = 2b + 24.5b

2b - 7a = 26.5b

The result is nonzero if and only if b is nonzero. Luckily we're told b is nonzero at the top of the problem. So we don't have any worries that (2b-7a) is zero.

Therefore, (2a+7b) + (2b-7a)i will be purely imaginary with a/b = -7/2

------------------

A concrete example:

Let a = -14 and b = 4

a/b = -14/4 = -7/2

(2-7i)(a+bi) = (2-7i)(-14+4i) = 0 + 106i which is purely imaginary.

I'll let you do the steps in expanding that out using the FOIL rule.

4 0

1 year ago

Which function is greater for<br> x < 2 (for all x values less than 2) ?

kodGreya [7K]

Answer:

Function 1

Step-by-step explanation:

8 0

3 years ago