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KonstantinChe [14]

2 years ago

14

24

1 answer:

Vaselesa [24]2 years ago

3 0

Answer:

1.5/2

Step-by-step explanation:

slope formula = y2-y1/ x2 - x1

point one (2,0)

point 2 (0, 1.5)

you dont really need to subtract anything because the intercepts, so the slope is 1.5/2

(slope or m = 1.5 - 0 / 2 - 0 )

x intercept = value of x when y is 0

y intercept = value of y when x is 0

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Flauer [41]

There are 30 treasures.

80% are found:
80% of 30 = 0.8 x 30 = 24
Bella found 24 of them

30 - 24 = 6
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2 years ago

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In-s [12.5K]

Answer: $1590cm^3$

Step-by-step explanation:

$V=\frac{3\sqrt{3} }{2}a^2h$

a = base

h = height

$V=\frac{3\sqrt{3} }{2}(6cm)^2(17cm)$

Solve the parentheses squared first.

$V=\frac{3\sqrt{3} }{2}(36cm^2)(17cm)$

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36/2=18

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Rounded to the nearest whole number... $1590cm^3$

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3 years ago

A baseball player had 4 hits in 8 games. At this rate, how many hits will the baseball player have in the next 28 games?

s2008m [1.1K]

14 hits

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8 0

3 years ago

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sammy [17]

Answer:

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Step-by-step explanation:

As sum of angles of the triangle is 180°

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3 years ago

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Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions

quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of $v_0$ and $F_2$ be the component normal to $v_0$ .

Since, $|v_0| = 1,$

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, $v_0 = \left$

From figure,

$|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3$

We know that the direction of $F_1$ is opposite of the direction of $v_0$ , so we have

$F_1 = -5\sqrt3 v_0$

$=-5\sqrt3 \left$

$= \left$

The unit vector in the direction normal to the plane, $v_1$ has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

$|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5$

∴ $F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>$

$=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

$= = F$

3 0

2 years ago