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2 years ago

Janiya received a 20% discount on the purchase of a $649 cell phone. What was the

Mathematics

1 answer:

avanturin [10]2 years ago

5 0

Answer:

$129.80

Step-by-step explanation:

Discount: .20 * 649 = 129.80

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Need help with this??? Please!!!

Radda [10]

Answer:

the first option

Step-by-step explanation:

(f-g)(x) simply means to subtract both expressions. really literally.

and we go through it power by power of x.

the highest power/exponent of x is 3 (x³). only f(x) has one.

so, -7x³ is the first part of f-g.

next is x².

11x² - 6x² = 5x², which is the second part of f-g.

next is x.

-8x - (-14x) = -8x + 14x = 6x, which is the third part of f-g.

next is x⁰ (in other words, no x, just a constant).

4 - (-3) = 4 + 3 = 7, which is the 4th part of f-g.

we have no x to the power of -1 or -2, so we have -3

0 - (-4x‐³) = 4x‐³, which is the last part of f-g.

so, it is clearly the first answer option.

3 0

2 years ago

LA plane is trying to travel 250 miles at a bearing of 20° E of S, however, it ends 230 miles away from the

Semmy [17]

Answer:

The wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ East of North with respect to the destination point.

Step-by-step explanation:

Let origin, O, br the starting point and point D be the destination at 250 miles at a bearing of 20° E of S, but due to wind let D' be the actual position of the plane at 230 miles away from the starting point in the direction of 35° E of South as shown in the figure.

So, we have |OD|=250 miles and |OD'|=230 miles.

Vector $\overrightarrow{DD'}$ is the displacement vector of the plane pushed by the wind.

From figure, the magnitude of the required displacement vector is

$|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)$

and the direction is $\alpha$ east of north as shown in the figure,

$\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)$

From the figure,

$|AB|=|OA-OB|$

$\Rightarrow |AB|=|OD\cos 20 ^{\circ}-OD'\cos 35 ^{\circ}|$

$\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|$

$\Rightarrow |AB|=45.52$ miles

Again, $|PQ|=|OP-OQ|$

$\Rightarrow |PQ|=|OD\sin 20 ^{\circ}-OD'\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=|250\sin 20 ^{\circ}-230\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=46.42$ miles

Now, from equations (i) and (ii), we have

$|DD'|=\sqrt{|45.52|^2+|46.42|^2}=65.01$ miles, and

$\tan \alpha=\frac{|46.42|}{|45.52|}$

$\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}$

Hence, the wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ E astof North with respect to the destination point.

5 0

2 years ago

Are 45/75 and 90/150 proportional?

lesya [120]

Yes they are
. 90/150= 3/5
so 45/75 and 3/5.

3 0

2 years ago

Read 2 more answers

Someone help?!.........

koban [17]

Answer: y = -2x + 16

Explanation:

Use point slope form:

(y1 - y) = m(x1 - x)

y1 = 6 and x1 = 5 and m = -2

6 - y = -2(5 - x)
6 - y = -10 + 2x
-y = 2x - 10 - 6
-y = 2x - 16
y = -2x + 16

5 0

2 years ago

Proportions and equations solve

kirza4 [7]

Y=9x because if you divided y by x all the numbers will equal to 9 which is proportional

7 0

2 years ago