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3 years ago

Find the distance between (2,4) and (12, 9) on a coordinate plane. round to the nearest tenth if necessary. *

1 answer:

4 0

Use pythagoras

d = √(Δx² + Δy²)
d = √(12-2)² + (9-4)²
d = √(10² + 5²)
d = √(100 + 25)
d = √125
d = 5√5
d = 5 × 2.236
d = 11.18

The nearest tenth is 11.2, so the distance is near to 11.2

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Can someone help me solve this? Pls and ty!

Zanzabum

Answer:

H. 95°

Step-by-step explanation:

m∠BGC is 10° smaller than m∠CGD, which means that m∠BGC is x° - 10°

m∠DGE is 25° more than m∠EGF, which means that m∠DGE is x° + 25°

The sum of all angles in a placement like this will always equal 360°

So we put all the measures of all angles on one side of thee equation and set it equal to 360*

x° + x° - 10° + 2x° + 25° + 2x° + 3x° + 30° = 360°

After combining like terms we get 9x° + 45° = 360°

After some algebra we get x = 35°

Now at the start we said that m∠DGE is x + 25°

Plug in 35° for x and we get 35° + 25°

The final answer is m∠DGE = 95°

Edit: Sorry if my answer looks a bit confusing. This is actually my first answer on Brainly so I'm quite new to this experience.

5 0

2 years ago

Read 2 more answers

The sum of three numbers in <br> g.p. is 21 and the sum of their squares is 189. find the numbers.

sashaice [31]

Let the three gp be a, ar and ar^2
a + ar + ar^2 = 21 => a(1 + r + r^2) = 21 . . . (1)
a^2 + a^2r^2 + a^2r^4 = 189 => a^2(1 + r^2 + r^4) = 189 . . . (2)
squaring (1) gives
a^2(1 + r + r^2)^2 = 441 . . . (3)
(3) ÷ (2) => (1 + r + r^2)^2 / (1 + r^2 + r^4) = 441/189 = 7/3
3(1 + r + r^2)^2 = 7(1 + r^2 + r^4)
3(r^4 + 2r^3 + 3r^2 + 2r + 1) = 7(1 + r^2 + r^4)
3r^4 + 6r^3 + 9r^2 + 6r + 3 = 7 + 7r^2 + 7r^4
4r^4 - 6r^3 - 2r^2 - 6r + 4 = 0
r = 1/2 or r = 2
From (1), a = 21/(1 + r + r^2)
When r = 2:
a = 21/(1 + 2 + 4) = 21/7 = 3
Therefore, the numbers are 3, 6 and 12.

3 0

3 years ago

PLEASE HURRY WITH THIS

Snezhnost [94]

Answer:

Step-by-step explanation:

100.5 in ^2

5 0

3 years ago

Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and

Lisa [10]

Answer:

Given :

ABC is a right triangle in which ∠ABC = 90°,

Also, Legs AB and CB are extended past point B to points D and E,

Such that,

$\angle EAC = \angle ACD = 90^{\circ}$

To prove :

$EB\times BD=AB\times BC$

Proof :

In triangles AEC and EBA,

∠EAC= ∠ABE ( right angles )

∠CEA = ∠AEB ( common angles )

By AA similarity postulate,

$\triangle AEC \sim \triangle EBA$ ,

Similarly,

$\triangle AEC \sim \triangle ABC$

$\implies\triangle EBA\sim \triangle ABC-----(1)$

Now, In triangles ADC and CBD,

∠ACD = ∠CBD ( right angles )

∠ADC= ∠BDC ( common angles )

By AA similarity postulate,

$\triangle ADC \sim \triangle CBD$ ,

Similarly,

$\triangle ADC \sim \triangle ABC$

$\implies \triangle CBD\sim \triangle ABC-----(2)$

From equations (1) and (2),

$\triangle EBA\sim \triangle CBD$

The corresponding sides of similar triangles are in same proportion,

$\frac{EB}{BC}=\frac{AB}{BD}$

$\implies EB\times BD=AB\times BC$

Hence, proved....

5 0

3 years ago

It’s confusing, would the answer be square root 7 over 4?

Colt1911 [192]

Answer:

$\tan U = 1$

Step-by-step explanation:

$\text{Apply Pythagorean theorem,}\\\\~~~~~~\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2\\\\\implies UT^2 = UV^2 +VT^2\\\\\implies UV^2 = UT^2 -VT^2\\\\\implies UV^2 = \left(7\sqrt 2 \right)^2 - 7^2\\\\\implies UV^2 = 49(2)-49\\\\\implies UV^2 = 49\\\\\implies UV = \sqrt{49} \\\\ \implies UV = 7$

$\text{Now,}\\\\~~~~~~~~\tan \theta = \dfrac{\text{Perpendicular}}{\text{Base}}\\\\\\\implies \tan U = \dfrac{7}{7}\\\\\\\implies \tan U = 1$

4 0

2 years ago