<img src="https://tex.z-dn.net/?f=%20%7Ba%20%7D%5E%7B2%7D%20%20%2B%208a%20%2B%2015%20%3D%200" id="TexFormula1" title=" {a }^{2}

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taurus [48]

3 years ago

+ 8a + 15 = 0" alt=" {a }^{2} + 8a + 15 = 0" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

NISA [10]3 years ago

8 0

Answer:

a = - 5, a = - 3

Step-by-step explanation:

Given

a² + 8a + 15 = 0 ← in standard form

(a + 3)(a + 5) = 0 ← in factored form

Equate each factor to zero and solve for a

a + 3 = 0 ⇒ a = - 3

a + 5 = 0 ⇒ a = - 5

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Leto [7]

7*2=14
14*3=42

Michelle is 42 years of age...

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3 years ago

55. REASONING How many solutions does the equation

schepotkina [342]

It will have 2 solutions! this is because a negative and a negative cancel each other out and will make a positive under the radical.

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3 years ago

Doug hits a baseball straight towards a 15 ft high fence that is 400 ft from home plate. The ball is hit 2.5ft above the ground

Scorpion4ik [409]

Answer:

$125.4\ \text{m/s}$

Step-by-step explanation:

u = Initial velocity of baseball

$\theta$ = Angle of hit = $30^{\circ}$

x = Displacement in x direction = 400 ft

y = Displacement in y direction = 15 ft

$y_0$ = Height of hit = 2.5 ft

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t = Time taken

Displacement in x direction

$x=u_xt\\\Rightarrow x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{400}{u\cos30^{\circ}}\\\Rightarrow t=\dfrac{400}{u\dfrac{\sqrt{3}}{2}}\\\Rightarrow t=\dfrac{800}{u\sqrt{3}}$

Displacement in y direction

$y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow y=y_0+u\sin\theta t+\dfrac{1}{2}a_yt^2\\\Rightarrow 15=2.5+u\sin30^{\circ}(\dfrac{800}{u\sqrt{3}})+\dfrac{1}{2}\times -32.2\times (\dfrac{800}{u\sqrt{3}})^2\\\Rightarrow \dfrac{400}{\sqrt{3}}-\dfrac{10304000}{3u^2}-12.5=0\\\Rightarrow 218.44=\dfrac{10304000}{3u^2}\\\Rightarrow u=\sqrt{\dfrac{10304000}{3\times218.44}}\\\Rightarrow u=125.4\ \text{m/s}$

The minimum initial velocity needed for the ball to clear the fence is $125.4\ \text{m/s}$

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3 years ago

<img src="https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BE%7D%7B%20%5Csqrt%7BR%20%7B%7D%5E%7B2%7D%20%2B%20W%20%7B%7D%5E%7B2%7D%20L%

AleksandrR [38]

Answer:

R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Step-by-step explanation:

Squaring both sides of equation:

I^2 = (ER)^2/(R^2 + (WL)^2)

<=>(ER)^2 = (I^2)*(R^2 + (WL)^2)

<=>(ER)^2 - (IR)^2 = (IWL)^2

<=> R^2(E^2 - I^2) = (IWL)^2

<=> R^2 = (IWL)^2/(E^2 - I^2)

<=> R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]

Hope this helps!

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3 years ago

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olya-2409 [2.1K]

Answer:

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Step-by-step explanation:

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2 years ago