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2 years ago

8

Solve for f. f + 15 < 6 or f + 12 > 16 _

1 answer:

Sergeeva-Olga [200]2 years ago

5 0

Step-by-step explanation:

1. f+15<6

= f<6-15

f < -9

2. f+12 >16

f >16-12

f > 4

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Alaina has $28 in her account. She wants to purchase a pair of shoes that costs $45. If Alaina makes the purchase, which integer

valentina_108 [34]

Answer:

The Integer is -17

Step-by-step explanation:

This is because you must subtract the cost of the shoes from her account

28 - 45 is equal to negative 17

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3 years ago

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2 years ago

Here is the histogram of a data distribution. All class widths are 1.

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2 years ago

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ICE Princess25 [194]

After staring at this for a few seconds, I think the equation

$-5\sqrt{10}\cdot\sqrt{15}=[1]\sqrt{[2]}$

is asking you to simplify the left hand side so that it has a form matching the right hand side, and [1] and [2] are placeholders. My guess is that they're supposed to be integers.

Notice that

$\sqrt{10}=\sqrt{5\cdot2}=\sqrt5\cdot\sqrt2$

$\sqrt{15}=\sqrt{5\cdot3}=\sqrt5\cdot\sqrt3$

$\implies\sqrt{10}\cdot\sqrt{15}=(\sqrt5)^2\cdot\sqrt2\cdot\sqrt3=5\sqrt{2\cdot3}=5\sqrt6$

Then multiplying by -5, we have

$-5\sqrt{10}\cdot\sqrt{15}=-25\sqrt6$

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2 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

3 years ago

Read 2 more answers