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2 years ago

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Caleb’s puppy weighs 2,250 grams. If the puppy weighed 600 grams at his last visit to the veterinarian’s office, what is the per

cent increase in the puppy’s weight rounded to the nearest whole number? Enter your answer in the box.

1 answer:

Fed [463]2 years ago

8 0

Answer:

$73\%$

Step-by-step explanation:

Given: Present weight of Caleb’s puppy = 2,250 grams

Puppy weighed 600 grams at his last visit to the veterinarian’s office.

To find: percent increase in the puppy’s weight rounded to the nearest whole number

Solution:

Increase in puppy's weight = 2250 - 600 = 1650 grams

Percent increase in the puppy’s weight =( Increase in puppy's weight/Present weight of Caleb’s puppy) × 100 = $\frac{1650}{2250}(100)=73.33 \%$

$73.33\%$ is rounded to $73\%$

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bezimeni [28]

M< 6 = m< 7 (vertical angles)
11x + 8 = <span>12x – 4
12x - 11x = 8 + 4
x = 12
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m< 6 = 11(12) + 8
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m< 6 = 140

m<4 = 180 - m<6
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m<4 = 40

answer

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Let x = 8.99999 . (a) Is x < 9 or is x = 9? x < 9 It is neither; x > 9 x = 9 x < 9 and x = 9 It cannot be determined

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Answer: idk

Step-by-step explanation:

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2 years ago

Circular tracts of land with diameters 900 meters, 700 meters and 600 meters are tangent to each other externally. There are hou

Rudik [331]

Answer:

$\alpha=69.28^o$

$\beta=61.26^o$

$\gamma=49.46^o$

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Step-by-step explanation:

<u>Triangle Solving</u>

If we had a triangle will its three sides of known length, we can solve for the rest of the parameters of the triangle, i.e. the area, perimeter and internal angles.

The three circles have diameters 900 m, 700 m and 600 m and are tangent to each other externally. The distances from their centers (where houses are located) are the sum of each pair of the radius of the circles. Thus, the sides of the triangle are

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$y=450+300=750$

$z=350+300=650$

The internal angles can be computed by using the cosine's law

$x^2=y^2+z^2-2yzcos\alpha$

$y^2=x^2+z^2-2xzcos\beta$

$z^2=x^2+y^2-2xycos\gamma$

Where \alpha, \beta and \gamma are the opposite angles to x, y and z respectively. Solving for each one of them:

$\displaystyle cos\alpha=\frac{y^2+z^2-x^2}{2yz}$

$\displaystyle cos\alpha=\frac{750^2+650^2-800^2}{2\cdot 750\cdot 650}$

$cos\alpha=0.3538$

$\alpha=69.28^o$

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$\displaystyle cos\beta=\frac{x^2+z^2-y^2}{2xz}$

$\displaystyle cos\beta=\frac{800^2+650^2-750^2}{2\cdot 800\cdot 650}$

$cos\beta=0.4808$

$\beta=61.26^o$

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$\displaystyle cos\gamma=\frac{x^2+y^2-z^2}{2xy}$

$\displaystyle cos\gamma=\frac{800^2+750^2-650^2}{2\cdot 800\cdot 750}$

$cos\gamma=0.65$

$\gamma=49.46^o$

The area can be found by

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denis-greek [22]

Answer:

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Step-by-step explanation:

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Hope this helps! :)

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