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3 years ago

What is 1 2/4 plus 9 5/10

Mathematics

2 answers:

german3 years ago

6 0

Answer,

Hope this helps :-)

Verdich [7]3 years ago

4 0

First we find the fraction in decimal form,

so they it will equal to 1.5 because 2/4 is 0.5 and 9.5 because 5/10 is .5 as well

Then we add and get a whole number which is 11

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The height of a triangle is 2 less than 5 times its base. If the base of the triangle is x feet, and the area of the triangle is

Semmy [17]

Answer:

A. 5x^2 − 2x − 24 = 0

Step-by-step explanation:

height = 5b-2

b =x

We can rewrite the height as

h = 5x-2

We know the formula for area of a triangle

A = 1/2 bh

A =1/2 x * (5x-2)

Distributing the x

A= 1/2 (5x^2 - 2x)

We know A =12

12 = 1/2 (5x^2 - 2x)

Multiply each side by 2

12*2 = 2*1/2 (5x^2 - 2x)

24 = 5x^2 - 2x

Subtract 24 from each side

24-24 = 5x^2 - 2x-24

0 = 5x^2 - 2x-24

6 0

3 years ago

Read 2 more answers

Can someone please help me with this. I'm not good with graphs

Vlad1618 [11]

Answer:

∠BCD = 34°

Step-by-step explanation:

∠FCE + ∠FCA + ∠ACB = 180° ( straight angle ), thus

∠FCE + 90 + 56 = 180, that is

∠FCE + 146 = 180 ( subtract 146 from both sides )

∠FCE = 34°, hence

∠BCD = ∠FCE = 34° ( vertical angles )

3 0

3 years ago

Find the exact value of the trigonometric expression.

Dmitrij [34]

$\bf \textit{Half-Angle Identities}
\\\\
sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}}
\qquad 
cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}\\\\
-------------------------------\\\\
\cfrac{1}{12}\cdot 2\implies \cfrac{1}{6}\qquad therefore\qquad \cfrac{\pi }{12}\cdot 2\implies \cfrac{\pi }{6}~thus~ \cfrac{\quad \frac{\pi }{6}\quad }{2}\implies \cfrac{\pi }{12}$

$\bf sec\left( \frac{\pi }{12} \right)\implies sec\left( \cfrac{\frac{\pi }{6}}{2} \right)\implies \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\impliedby \textit{now, let's do the bottom}
\\\\\\
cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( \frac{\pi }{6} \right)}{2}}\implies \pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies \pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}}$

$\bf \pm \sqrt{\cfrac{2+\sqrt{3}}{4}}\implies \pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\implies \pm \cfrac{\sqrt{2+\sqrt{3}}}{2}
\\\\\\
therefore\qquad \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\implies \cfrac{2}{\sqrt{2+\sqrt{3}}}$

which simplifies thus far to

$\bf \cfrac{2}{\sqrt{2+\sqrt{3}}}\cdot \cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\cdot \stackrel{conjugate}{\cfrac{2-\sqrt{3}}{2-\sqrt{3}}}
\\\\\\
\cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{2^2-(\sqrt{3})^2}\implies \cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{1}$

$\bf 2\sqrt{2+\sqrt{3}}(2-\sqrt{3})\impliedby \stackrel{\textit{keep in mind that}}{(2-\sqrt{3})=(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}})}
\\\\\\
2\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}
\\\\\\
2\sqrt{(2+\sqrt{3})(2-\sqrt{3})\cdot (2-\sqrt{3})}
\\\\\\
2\sqrt{[2^2-(\sqrt{3})^2]\cdot (2-\sqrt{3})}\implies 2\sqrt{1(2-\sqrt{3})}
\\\\\\
2\sqrt{2-\sqrt{3}}$

7 0

4 years ago

Has anyone know the answer

dem82 [27]

Answer:

The answer is A

The cost of 5 tickets is $40

4 0

1 year ago

Read 2 more answers

Solve for p. -1 + 5p = -10 + 4p

sdas [7]

Answer:

p = - 9

Step-by-step explanation:

Given

- 1 + 5p = - 10 + 4p ( subtract 4p from both sides )

- 1 + p = - 10 ( add 1 to both sides )

p = - 9

5 0

3 years ago