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3 years ago

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Teo rides his bike in a straight line from his location perpendicular to path A, and Luke rides his bike in a straight line from

his location, perpendicular to path B. What are the coordinates of the point where they meet?

1 answer:

ollegr [7]3 years ago

6 0

Answer:

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Hi guys ummm what 28 multiplied by 4

kodGreya [7K]

Answer:

112 :D hope this helps

Step-by-step explanation:

112

4 0

3 years ago

Read 2 more answers

Find the value of x of this question

mel-nik [20]

<h3><u>Let's</u><u> </u><u>understand</u><u> </u><u>the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°
So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle
So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

8 0

3 years ago

I have $225.00 in my account. I can spend al but 20% of that on some new equipment. What is the most that I can afford to spend

Setler79 [48]

You can "spend all but 20%" so that means you can spend up to 80% of your money (since 20%+80% = 100%). Apply 80% to the value 225

80% of 225 = (80/100)*225 = 0.80*225 = 180

The most you can afford to spend is $180

3 0

4 years ago

T= 1,-1 u= 2, -4

Monica [59]

Answer: 3.16 units (Option C)

Step-by-step explanation:

The key to this problem is to use the distance formula, which is:

Distance = $\sqrt{(x₁-x₂)²+(y₁-y₂)²}$

The first point, T = (x₁,y₁), and the second point, U = (x₂,y₂).

Plugging the two points into the equation, we get:

Distance = $\sqrt{(1-2)^{2}+(-1-(-4))^{2} }$

The values within the parenthesis are subtracted:

Distance = $\sqrt{(-1)^{2}+3^{2} }$

The values are then squared:

Distance = $\sqrt{1+9}$

Finally, they are added together:

Distance = $\sqrt{10}$

$\sqrt{10}$ can be approximated as 3.16, so the distance between the two points is 3.16 units.

7 0

3 years ago

Need help<br><br><br> with mathhhhhhhhhh

tia_tia [17]

Just plug in your x and solve like normal

-4< 4(1) - 8 < 12
-4< -4 < 12

now solve both sides by moving the -4 out by adding + 4 to all sides

0 < x < 16

when read this says 0 is less than X and X is less than positive 16.

the other 3 are solved the same.

7 0

3 years ago