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2 years ago

Fix the mistake X - 45 = .27

Mathematics

2 answers:

s344n2d4d5 [400]2 years ago

8 0

Answer:

It is 45.27

Step-by-step explanation:

blondinia [14]2 years ago

5 0

I think that it’s 45.27

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If 7x+2a=3x+5a then x is equivalent to what

solniwko [45]

7x+2a=3x+5a

7x-3x = 5a -2a

4x = 3a

x = 3a/4

4 0

2 years ago

Read 2 more answers

An isosceles triangle has a vertical angle of 108 and a base 20 cm long. Solve for the base angles in calculated altitude.

xeze [42]

Answer:

The base angles are 36° each

Step-by-step explanation:

An isosceles triangle is a triangle that has two of its sides and base angles to be equal.

If an isosceles triangle has a vertical angle of 108°, then the base angles will share the remaining angle in the triangle equally.

Since the sum of angles in a triangle is 180°

The remaining angle in the triangle = 180°-108°

= 72°

Since the base angles are equal, this remaining two angles will share angle 72° equally.

The base angles = 72°/2

The base angles = 36°

This shows that the base angles are 36° each.

8 0

2 years ago

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$