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2 years ago

∠A and \angle B∠B are vertical angles. If m\angle A=(7x-8)

Mathematics

1 answer:

Anika [276]2 years ago

5 0

Answer:

x=25

Step-by-step explanation:

We know vertical angles are equal

<A = <B

7x-8 = 6x+17

Subtract 6x from each side

7x-6x-8 = 6x-6x+17

x-8 = 17

Add 8 to each side

x-8+8 = 17+8

x=25

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The answer is a cause Alan

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Natalie had 64 beads. She used 24 beads to make a bracelet. She used 7/8 of the remaining beads to make a necklace. How many bea

Eddi Din [679]

Answer:

Natalie have left with 5 beads.

Step-by-step explanation:

Given:

Natalie had 64 beads. She used 24 beads to make a bracelet. She used 7/8 of the remaining beads to make a necklace.

Now, to find the beads does Natalie have left.

Total beads Natalie had = 64 beads.

And beads she used to make a bracelet = 24 beads.

So, the remaining beads are = $64-24=40\ beads.$

Now, the beads she used to make a necklace = $40 of \frac{7}{8}$

= $40\times \frac{7}{8}$

= $\frac{280}{8}$

= $35\ beads$

Hence, the beads Natalie have left = remaining beads - beads used to make necklace.

$The\ beads\ Natalie\ have\ left\ =\ 40-35$

$The\ beads\ Natalie\ have\ left\ =\ 5\ beads.$

Therefore, Natalie have left with 5 beads.

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2 years ago

In DCAB, if ÐB = 38', What is ÐC?

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2 years ago

Weight: How many grams does a 5 lb 15 oz roast weigh?The roast weighsgrams.

natulia [17]

Since 1kg = 2.20 lb,

$\frac{1\operatorname{kg}}{2.20\text{ lb}}$

Since 1kg = 1000g

$\frac{1000\text{ g}}{1\operatorname{kg}}$

So for covering pounds to gram by the following factor

$\frac{1\operatorname{kg}}{2.20\text{ lb}}\frac{1000\text{ g}}{1\operatorname{kg}}$

Since 1oz = 28.3g

$\frac{28.3\text{ g}}{1\text{ oz}}$

Now

$5\text{ lb}\frac{1000\text{ g}}{2.20\text{ lb}}+15\text{ oz}\frac{28.3g}{1\text{ oz}}$

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2697.22 gr

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11 months ago

A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th

Yuri [45]

Answer:

$15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

$y(0)=0$ , $y'(0)=0$

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: $F_{r} (t)$ that correspond to the force of resistance on the mass by the action of the spring and $F(t)$ that is an external force with unknown direction (that does not specify in the enounce).

For determinate $F_{r} (t)$ we can use Hooke's Law given by the formula $F_{r} (t) = k y(t)$ where $k$ correspond to the elastic constant of the spring and $y(t)$ correspond to the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

$k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m} = 441.45 \frac{N}{m}$

Now we apply Newton's 2nd Law and obtaint that...

$F_{r} (t)$ ± $F(t)$ = $ma(t)$

$F_{r} (t) = ky(t) = 441.45y(t)$

$F(t) = 170 cos(5t)$

$m = 15 kg$

$a(t) = \frac{d^{2}y(t)}{dt^{2} }$

Finally... $15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

We know from the problem that there's not initial displacement and initial velocity, so... $y(0)=0$ and $y'(0)=0$

Finally the Initial Value Problem that models the situation describe by the problem is

$\left \{ 15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.$

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2 years ago