All categories

2 years ago

Do the ratios 18 7 and 2 1 form a proportion?

Mathematics

1 answer:

deff fn [24]2 years ago

5 0

Hi!

There are a few ways we can do this. The first one is this:

The ratios 18:7 and 2:1 can be represented as fractions: $\frac{18}{7}$ and $\frac{2}{1}$

To see if they are proportions, we can put the fractions in their simplest form and see if they are equal. However, they are already simplified, and we can see that they are not equal.

Therefore they are not proportions.

The second way we can see is to use cross products. Once again we'll put them as fractions, except we'll set them side by side, like so:

$\frac{18}{7}$ $\frac{2}{1}$

To see if they are proportions, we simply multiply the diagonals. These would be 18 x 1 and 7 x 2 which equals 18 and 14. These are not equal.

Therefore they are not proportions.

Hope this helps! Have a great day! Let me know if you need any other help :D

You might be interested in

Make a equation for the scatter plot line

aalyn [17]

Answer:

y = -7.2x + 540

Step-by-step explanation:

to find the slope I used the x-and-y intercepts: (0, 540) and (75, 0)

slope = 540/-75 or -7.2

5 0

3 years ago

Q bisects PR, PQ=3y, and PR=42. Find y and QR.

Vedmedyk [2.9K]

|PR| = |PQ| + |QR|; |PQ| = |QR| conclusion |PR| = 2|PQ|

|PQ| = 3y; |PR| = 42; |QR|=?

subtitute

42 = 2(3y)
6y = 42 |divide both sides by 6
y = 7

|QR| = 3(7) = 21

3 0

3 years ago

Solve the inequality -1/3h ≥ 7 A. h ≥ –21 B. h ≤ 7 1/3 C. h ≤ –21 D. h ≥ -2 1/3

JulsSmile [24]

The Answer Is C Fam...................

8 0

3 years ago

Read 2 more answers

Please show your work this question what made you come to the conclusion. Thank you

AveGali [126]

Answer:

B the range, the x- and y-intercept

Step-by-step explanation:

the domain stays the same : all values of x are possible out of the interval (-infinity, +infinity).

but the range changes, as for the original function y could only have positive values - even for negative x.

the new function has a first term (with b) that can get very small for negative x, and then a subtraction of 2 makes the result negative.

the y-intercept (x=0) of the original function is simply y=1, as b⁰=1.

the y-intercept of the new function is definitely different, because the first term 3×(b¹) is larger than 3, because b is larger than 1. and a subtraction of 2 leads to a result larger than 1, which is different to 1.

the original function has no x-intercept (y=0), as this would happen only for x = -infinity. and that is not a valid value.

the new function has an x-intercept, because the y-values (range) go from negative to positive numbers. any continuous function like this must therefore have an x-intercept (again, y = the function result = 0)

$3 {b}^{x + 1} = 2$

${b}^{x + 1} = 2 \div 3$

$log_{b}(2 \div 3) = x + 1$

$x = log_{b}(2 \div 3) - 1$

8 0

3 years ago

Which of the following is not one of the 8th roots of unity?

Anika [276]

Answer:

1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1. Since $z=1=1+0\cdot i,$ then

$Rez=1,\\ \\Im z=0$

and

$|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.$

This gives you $\varphi=0.$

Thus,

$z=1\cdot(\cos 0+i\sin 0).$

2. The 8th roots can be calculated using following formula:

$\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.$

Now

at k=0, $z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;$

at k=1, $z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=2, $z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;$

at k=3, $z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=4, $z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;$

at k=5, $z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

at k=6, $z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;$

at k=7, $z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

The 8th roots are

$\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.$

Option C is icncorrect.

5 0

3 years ago