All categories

3 years ago

What is one factor of 30x2 − 33x + 3?

Mathematics

1 answer:

olganol [36]3 years ago

8 0

Answer:

10x - 1

Step-by-step explanation:

30x^2 − 33x + 3 =

= 3(10x^2 - 11x + 1)

= 3(10x - 1)(x - 1)

Answer: 10x - 1

You might be interested in

If the radius of a circle is 8 feet, please find:

Alexandra [31]

Step-by-step explanation:

Radius (r) = 8 feet

$Diameter(d) = 2 \times radius \\= 2 \times 8 = 16 \: feet \\ \\ circumference \\ c = \pi \: d = 3.14 \times 16 = 50.24 \: feet \\ \\ area = \pi {r}^{2} \\ = 3.14 \times {8}^{2} \\ = 3.14 \times 64 \\ = 200.96 \: {feet}^{2}$

5 0

4 years ago

Helpppp pleaseeeeeeeeeeeeeee

Vesna [10]

Answer:

Find below the calculations of the two areas, each with two methods. The results are:

Upper triangle:

$Area=5000\sqrt{3}units^2$

Lower triangle:

$Area=14,530m^2$

Explanation:

A) Method 1

When you are not given the height, but you are given two sides and the included angle between the two sides, you can use this formula:

$Area=side_1\times side_2\times sin(\alpha)$

Where, $\alpha$ is the measure of the included angle.

1. Upper triangle:

$side_1=200units\\ \\ side_2=100units\\ \\ \alpha =60\º\\ \\ Area=200units\times 100units\times sin(60\º)/2\\ \\ Area=5000\sqrt{3}units^2$

2. Lower triangle:

$side_1=231m\\ \\ side_2=150m\\ \\ \alpha =123\º\\ \\ Area=231m\times 150m\times sin(123\º)/2\\ \\ Area=14,529.96m^2\approx14,530m^2$

B) Method 2

You can find the height of the triangle using trigonometric properties, and then use the very well known formula:

$Area=(1/2)\times base\times height$

Use it for both triangles.

3. Upper triangle:

The trigonometric ratio that you can use is:

$sine(\alpha)=opposite\text{ }leg/hypotenuse$

Notice the height is the opposite leg to the angle of 60º, and the side that measures 100 units is the hypotenuse of that right triangle. Then:

$sin(60\º)=height/100units\\ \\ height=sin(60\º)\times100units\\ \\ height=50\sqrt{3}units$

$Area=(1/2)\times base\times height=(1/2)\times 200units\times 50\sqrt{3}units=5,000\sqrt{3}units^2$

3. Lower triangle:

$sin(180\º-123\º)=height/231m\\ \\ height=sin(57\º)\times 231m\\ \\ height=193.7329m^2$

$Area=(1/2)\times base\times height=(1/2)\times 150m\times 193.7329m^2\\\\ Area=14,529.96m^2\approx 14,530m^2$

5 0

3 years ago

.. Which of the following are the coordinates of the vertices of the following square with sides of length a?

atroni [7]

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Step-by-step explanation:

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

To find the sides of a square, let us use the distance formula,

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } S T=\sqrt{(a-0)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } T W=\sqrt{(a-a)^{2}+(0-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,a), T(a,a), W(a,0)$ has sides of length a.

Option B: $O(0,0), S(0,a), T(2a,2a), W(a,0)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text {Length } S T=\sqrt{(2 a-0)^{2}+(2 a-a)^{2}}=\sqrt{5 a^{2}}=a \sqrt{5}\\&\text {Length } T W=\sqrt{(a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{2 a^{2}}=a \sqrt{2}\\&\text {Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

This is not a square because the lengths are not equal.

Option C: $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length OS }=\sqrt{(0-0)^{2}+(2 a-0)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } S T=\sqrt{(2 a-0)^{2}+(2 a-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } T W=\sqrt{(2 a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } O W=\sqrt{(2 a-0)^{2}+(0-0)^{2}}=\sqrt{4 a^{2}}=2 a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$ has sides of length 2a.

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length OS }=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } S T=\sqrt{(a-a)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } T W=\sqrt{(0-a)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } O W=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

Thus, the square with vertices $O(0,0), S(a,0), T(a,a), W(0,a)$ has sides of length a.

Thus, the correct answers are option a and option d.

8 0

3 years ago

. After a blizzard, 18 inches of snow sat on Maggie's driveway. She measured the snow each

den301095 [7]

Answer:

13.5

Step-by-step explanation:

answer 18-4-0.25

struggling dub deth duck

3 0

3 years ago

Who is the father of mathematics

gayaneshka [121]

Hi there!

I believe the father or modern mathematics is René Descartes.

Hope this helped!~

7 0

4 years ago

Read 2 more answers