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3 years ago

I need help on this please help me

Mathematics

2 answers:

lbvjy [14]3 years ago

4 0

Answer:

(0,-4-4√6) and (0,-4+4√6)

Step-by-step explanation:

To determine the coordinates as Jon wants to place it 14 points away

√10^2+(a+4)^2 = 14

100+(a+4)^2 = 196

(a+4)^2 = 96

a+4 = ±√96

a+4 = ±4√6

a= -4±4√6

Oxana [17]3 years ago

3 0

On positive 3 on the x axis it

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Simplify. 2x+7y+3x−4y please help

Mariulka [41]

Answer:

Step-by-step explanation:

$2x+7y+3x-4y\\\\\mathrm{Group\:like\:terms}\\=2x+3x+7y-4y\\\\\mathrm{Add\:similar\:elements:}\:2x+3x=5x\\=5x+7y-4y\\\\\mathrm{Add\:similar\:elements:}\:7y-4y=3y\\= 5x+3y$

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3 years ago

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As a ship approaches the dock, it forms a 70 angle between the dock and the lighthouse. At the lighthouse, an 80 angle is formed

Trava [24]

Answer:

The distance from the ship to the dock is approximately 5.24 miles

Step-by-step explanation:

From the parameters given in the question, we have;

The angle formed between the dock and the lighthouse = 70°

The angle formed between the dock and the lighthouse at the ship = 80°

The distance between dock and the lighthouse = 5 miles (From a similar question online)

By sine rule, we have;

$\dfrac{a}{sin(A)} = \dfrac{b}{sin(B)} = \dfrac{c}{sin(C)}$

Therefore, we have;

$\dfrac{5}{sin(70^{\circ})} = \dfrac{The \ distance \ from \ the \ ship \ to \ the \ dock}{sin(80^{\circ})}$

$\therefore The \ distance \ from \ the \ ship \ to \ the \ dock = sin(80^{\circ}) \times \dfrac{5}{sin(70^{\circ})}$

$sin(80^{\circ}) \times \dfrac{5}{sin(70^{\circ})} \approx 5.24 \ mi$

Therefore;

The distance from the ship to the dock ≈ 5.24 miles

7 0

3 years ago

Select all the equations that are equivalent to the equation 3x – 4 = 5

erik [133]

A B E

They all equal x = 3 once solved.
C = 6 and D = 9 so those are not correct.

7 0

3 years ago

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Algebra 1 need help ASAP

White raven [17]

Answer:

Step 1: 12-2a=3a-18

Step 2: 12-5a=-18

Step 3: -5a=-30

Step 4: a=6

I hope this helps!

pls ❤ and give brainliest pls

7 0

3 years ago

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marcela tardo dos horas menos que roberto en recorrer una distancia en bicicleta, SI el producto de los dos tiempos es de 48 hor

Kaylis [27]

Answer:

Marcela tardó 6 horas y Roberto tardó 8 horas.

Step-by-step explanation:

Dado que Marcela tardó dos horas menos que Roberto en recorrer una distancia en bicicleta, y sabiendo que el producto de los dos tiempos es de 48 horas, para determinar cuánto tardó cada uno se debe realizar el siguiente cálculo:

M = X

R = X + 2

X x (X + 2) = 48

8 x 6 = 48

Por lo tanto, Marcela tardó 6 horas y Roberto tardó 8 horas.

8 0

3 years ago