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4 years ago

12

A adult blinks about 450 times every 30 minutes. a 12 year old blinks about 150 times every 15 minutes. how many more times does

a adult blink every hour than a 12 year old

1 answer:

ahrayia [7]4 years ago

4 0

An adult blinks 300 times more in an hour

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Given: circle k(O), m∠P=95°, m∠J=110°, m∠LK=125°<br> Find: m∠PJ

IRINA_888 [86]

Answer:

The measure of the arc PJ is $75\°$

Step-by-step explanation:

step 1

Find the measure of angle L

we know that

In a inscribed quadrilateral opposite angles are supplementary

so

$m$

we have

$m$

substitute

$m$

$m$

step 2

Find the measure of arc KJ

we know that

The inscribed angle measures half that of the arc comprising

so

$m$

substitute the values

$95\°=\frac{1}{2}(125\°+arc\ KJ)$

$190\°=(125\°+arc\ KJ)$

$arc\ KJ=190\°-125\°=65\°$

step 3

Find the measure of arc PJ

we know that

The inscribed angle measures half that of the arc comprising

so

$m$

substitute the values

$70\°=\frac{1}{2}(65\°+arc\ PJ)$

$140\°=(65\°+arc\ PJ)$

$arc\ PJ=140\°-65\°=75\°$

5 0

3 years ago

Please help me with this

LekaFEV [45]

Answer: can't add an answer, but for the first all 3 angles add up to 180 in a triangle for the second, plug 5 in and also check if all three add up to 180

Step-by-step explanation:

8 0

3 years ago

Two numbers that multiply to -100 and add to 15

Whitepunk [10]

Answer:

-5 and 20

Step-by-step explanation:

-5 + 20 = 15

-5 * 20 = -100

3 0

3 years ago

Help me please??!!!

kaheart [24]

Answer:

its the second 1

Step-by-step explanation:

i had last week and got a 88.87%

6 0

3 years ago

Trigonometry pile up. Anyone mind helping me out with this one?

S_A_V [24]

$cos34^o=\frac{8}{a};\ cos34^o\approx0.829\\\\\frac{8}{a}=0.829\to a\approx9.65\\\\9.65-2.5+3.6=10.75\\\\cos11^o=\frac{b}{10.75};\ cos11^o\approx0.981\\\\\frac{b}{10.75}=0.981\to b\approx10.55\\\\10.55-2.1+2.9=11.35\\\\c^2+3.8^2=11.35^2\\\\c^2=11.35^2-3.8^2\\\\c\approx10.69$

$10.69-3.2=7.49\\\\tan42^o=\frac{d}{7.49};\ tan42^o\approx0.9\\\\\frac{d}{7.49}=0.9\to d\approx6.74\\\\sin37^o=\frac{6.74}{e};\ sin37^o\approx0.602\\\\\frac{6.74}{e}=0.602\to e\approx11.2\\\\11.2+4.3=15.5\\\\sin53^o=\frac{f}{15.5};\ sin53^o\approx0.8\\\\\frac{f}{15.5}=0.8\to f\approx12.4$

$12.4-2.2=10.2\\\\g^2=10.2^2+1.7^2\to g\approx10.34\\\\cos21^o=\frac{10.34}{h};\ cos21^o\approx0.934\\\\\frac{10.34}{h}=0.934\to h=11.07\\\\11.07+1.7=12.77\\\\sin71^o=\frac{x}{12.77};\ sin71^o\approx0.946\\\\\frac{x}{12.77}=0.946\to x\approx12.1\ (cm)\leftarrow Answer$

4 0

3 years ago