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3 years ago

What's the answer to 16-12+12x17

Mathematics

2 answers:

daser333 [38]3 years ago

4 0

16-12+12x17= 208 u should use a calcuator

Sav [38]3 years ago

4 0

16-12+12×17,according to pemdas
P
E
M
D
A
S
the answer is 208

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Lnes made9 necklaces with 60 beads on each necklaces. How many beads did she use in all

Gala2k [10]

Made 9 necklaces.
60 beads per necklace.
How many beads for all the 9 necklaces?

60 × 9 = 540

9 necklaces = 540 beads<span />

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2 years ago

Find the slope intercept form of the line passing through (3,9) and (0,3).

ANEK [815]

Answer:

y = 2x +3

Step-by-step explanation:

To find the slope we do (Y1 - Y2)/(X1-X2) aka rise over run

So, m = (9 - 3) / (3 - 0)

m = 2

Now we have y = 2x +b

Notice one of our points is (0,3)

Since the x value of that point is 0, it is the y intercept.

In the slope intercept form b = y-intercept

Therefore b = 3

So, y = 2x + 3

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3 years ago

How to evaluate the expression 0.3(3-0.5)

scoray [572]

0.3(3 - 0.5)
= 0.3(2.5)
= 0.75

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3 years ago

A migrating whale is traveling at an average speed of 5.4 kilometers per hour. How far will the whale travel in 18 hours at this

zaharov [31]

97.2 miles just multiply

5.4
18
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3 years ago

Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -

GaryK [48]

Consider all parabolas:

$y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.$

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

$y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.$

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

$y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.$

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

$y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.$

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

$y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.$

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

$y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.$

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

3 0

3 years ago