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2 years ago

WILL MARK BRAINLIEST!!!

Mathematics

2 answers:

Art [367]2 years ago

8 0

Answer:

<em>I think the best answer will be is</em><em> C. 1.3 Good Luck!</em>

professor190 [17]2 years ago

8 0

Answer:

Step-by-step explanation:

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The quarterback of a football team releases a pass at a height of 6 feet above the playing field, and the football is caught by

motikmotik

Answer:

t = 25.1seconds

Step-by-step explanation:

X(t)= 16cos75°t

Y(t)= 6 + (Vosin75°t - 16t^2)

Converting 78 yards to feet :

1 yard = 3 feet

78feet =?

78×3=234 feet

Y(t)= number of feet above the ground at t seconds

VoCos75° t = 234

t = 234/(VoCos75°)

At this time,4 = y(234/VoCos75°)

4 = 6 + VoSin75°(234/VoCos75) - 16(234/VoCos75°)^2

Vo= 140.66ft/s

Time,t= 234/(140.66× Cos75°)

t = 25.087seconds

t= 25.1 seconds (1 decimal place)

8 0

2 years ago

Please Help me I am struggling a lot and I would be very grateful!!!

yKpoI14uk [10]

The answer would be iii: 10-15%
the way to find this is to divide 43 by 381 to get 11.2% which falls in the rage of 10-15%

5 0

3 years ago

according to the fundamental theorem of algebra , how many zeros does the polynomial below have x^4 +5x^3+10x^2+20x+24

Vinil7 [7]

Answer:

Step-by-step explanation:

the Fundamental Theorem of Algebra states that for any polynomial of degree n, there are n roots, some of which may be complex

The polynomial shown is of degree 4 ( highest exponent of x )

Hence the polynomial has 4 roots/ zeros

8 0

2 years ago

Read 2 more answers

Rationalise the denominator of:<br>1/(√3 + √5 - √2)

Paul [167]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} }$

can be re-arranged as

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} - \sqrt{2} + \sqrt{5} }$

$\rm \: = \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} }$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} } \times \dfrac{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }$

We know,

$\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using this, we get

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ {( \sqrt{3} - \sqrt{2} )}^{2} - {( \sqrt{5}) }^{2} }$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{3 + 2 - 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{5 - 2 \sqrt{6} - 5}$

$\rm \: = \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ - 2 \sqrt{6}}$

$\rm \: = \: \dfrac{ - ( - \sqrt{3} + \sqrt{2} + \sqrt{5}) }{ - 2 \sqrt{6}}$

$\rm \: = \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}}$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}} \times \dfrac{ \sqrt{6} }{ \sqrt{6} }$

$\rm \: = \: \dfrac{- \sqrt{18} + \sqrt{12} + \sqrt{30}}{2 \times 6}$

$\rm \: = \: \dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}$

$\rm \: = \: \dfrac{- 3\sqrt{2} + 2 \sqrt{3} + \sqrt{30}}{12}$

Hence,

$\boxed{\tt{ \rm \dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} } =\dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}}}$

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<h3><u>More Identities to </u><u>know:</u></h3>

$\purple{\boxed{\tt{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}}}}$

$\purple{\boxed{\tt{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}}}$

$\purple{\boxed{\tt{ {(x + y)}^{3} = {x}^{3} + 3xy(x + y) + {y}^{3}}}}$

$\purple{\boxed{\tt{ {(x - y)}^{3} = {x}^{3} - 3xy(x - y) - {y}^{3}}}}$

$\pink{\boxed{\tt{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}}$

$\pink{\boxed{\tt{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}}}$

6 0

3 years ago

A quadratic equation of the form 0 = ax2 + bx + c has one real number solution. Which could be the equation?

Fynjy0 [20]

Answer:

The equation showing this situation is $D=b^2-4ac=0$

Step-by-step explanation:

Given : A quadratic equation of the form $ax^2 + bx + c=0$ has one real number solution.

To find : Which could be the equation?

Solution :

A quadratic equation in form $ax^2+bx+c=0$ has a solution $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ called a quadratic formula in which the roots are one real,two real or no real is determine by discriminant factor.

Discriminant is defined as to determine the number of roots in a quadratic equitation has following rules :

1) If $D=b^2-4ac>0$ there are two real roots.

2) If $D=b^2-4ac=0$ there are one real roots.

3) If $D=b^2-4ac$ there are no real roots.

According to question,

A quadratic equation of the form $ax^2 + bx + c=0$ has one real number solution.

So, The equation showing this situation is $D=b^2-4ac=0$

6 0

3 years ago