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3 years ago

What is the measure of arc FH?

Mathematics

1 answer:

dmitriy555 [2]3 years ago

5 0

Answer:

140

Step-by-step explanation:

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larry mitchell invested part of his $27,000 advance at 2% annual simple interest and the rest at 6% annual simple interest. If h

Studentka2010 [4]

X=$4,000 INVESTED5%.

28,000-4,000=$24,000 INVESTED 3%.

Step-by-step explanation:

PROOF:

.05*4,000+.03*24,000=920

200+720=920

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4 0

3 years ago

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What is 160 bananas; 20% decrease??

inessss [21]

Answer:

128

Step-by-step explanation:

160 x .20 = 32

160 - 32 = 128

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4 years ago

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ELEN [110]

1-1/6-1/9=3/18

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3 years ago

It is 20 +a = 29<br> a= ?

emmasim [6.3K]

Answer:

9 is the answer :)

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4 years ago

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Help me find this asap

sammy [17]

Answer:

DQ = $\frac{5}{(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}-\sqrt{x+h})}$

Step-by-step explanation:

Given function is f(x) = $\frac{5}{2-\sqrt{x}}$

f(x + h) = $\frac{5}{2-\sqrt{x+h} }$

Therefore, indicated difference quotient will be,

DQ = $\frac{f(x+h)-f(x)}{h}$

Now we substitute the values in the difference quotient,

DQ = $\frac{\frac{5}{2-\sqrt{x+h} }-\frac{5}{2-\sqrt{x}}}{h}$

= $\frac{5(2-\sqrt{x})-5(2-\sqrt{x+h})}{h(2-\sqrt{x})(2-\sqrt{x+h})}$

= $\frac{10-5\sqrt{x}-10+5\sqrt{x+h}}{h(2-\sqrt{x})(2-\sqrt{x+h})}$

= $\frac{-5\sqrt{x}+5\sqrt{x+h}}{h(2-\sqrt{x})(2-\sqrt{x+h})}$

= $\frac{5(-\sqrt{x}+\sqrt{x+h})}{h(2-\sqrt{x})(2-\sqrt{x+h})}$

= $\frac{5(-\sqrt{x}+\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}{h(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}$

= $\frac{5[(x+h)-x]}{h(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}$

= $\frac{5h}{h(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}$

= $\frac{5}{(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}$

6 0

3 years ago