MATHEMATICS HIGH SCHOOL

Use an algebraic equation to solve the problem. The sides of a triangle are in the ratio 3 : 4 : 5. What is the length of each side if the perimeter of the triangle is 90 cm? A. 10.5 cm, 11.5 cm, and 12.5 cm

B. 22.5 cm, 30 cm, and 37.5 cm

C. 7.5 cm, 11.5 cm, and 32.1 cm

D. 19.3 cm, 25.7 cm, and 32.1 cm

Answers

Answer 1

Answer:

Answer:

The answer is the option B

$22.5\ cm, 30\ cm, 37.5\ cm$

Step-by-step explanation:

Let

a,b,and c the length sides of the triangle

we know that

$a+b+c=90$ ------> equation A

$(a)/(b)= (3)/(4)$

$a=(3)/(4)b$ -----> equation B

$(b)/(c)= (4)/(5)$

$c=(5)/(4)b$ -----> equation C

Substitute the equation B and equation C in the equation A

$(3)/(4)b+b+(5)/(4)b=90$

solve for b

$3b=90$

$b=30\ cm$

Substitute the value of b in the equation B

$a=(3)/(4)(30)=22.5\ cm$

Substitute the value of b in the equation C

$c=(5)/(4)(30)=37.5\ cm$

Answer 2

Answer:

Ratio units = 3+4+5 = 12

perimeter = 90 (a+b+c)

1 unit value, 90/12 = 7.5

So, lengths will be 3(7.5), 4(7.5), 5(7.5)

OPTION B IS YOUR ANSWER

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Given cos alpha = 8/17, alpha in quadrant IV, and sin beta = -24/25, beta in quadrant III, find sin(alpha-beta)

Answers

Given $\cos\alpha=(8)/(17)$ , $\alpha$ is in Quadrant IV, $\sin\beta=-(24)/(25)$ , and $\beta$ is in Quadrant III, find $\sin(\alpha-\beta)$

We can use the angle subtraction formula of sine to answer this question.

$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$

We already know that $\cos\alpha=(8)/(17)$ .

We can use the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$ to find $\sin\alpha$ .

$\sin^2\alpha+((8)/(17))^2=1 \n \sin^2\alpha+(64)/(289)=1 \n \sin^2\alpha=(225)/(289) \n \n\sin\alpha=\pm(15)/(17)$

Since $\alpha$ is in Quadrant IV, and sine is represented as y value on the unit circle, we must assume the negative value $\sin\alpha=-(15)/(17)$ .

As similar process is then done with $\sin\beta=-(24)/(25)$ .

$(-(24)/(25))^2+\cos^2\beta=1 \n (576)/(625)+\cos^2\beta=1 \n \cos^2\beta=(49)/(625) \n \n\cos\beta=\pm(7)/(25)$

And since $\beta$ is in Quadrant III, and cosine in represented as x value on the unit cercle, we must assume the negative value $\cos\beta=-(7)/(25)$ .

Now we can fill in our angle subtraction formula!

$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta \n\n \sin(\alpha-\beta)=(-(15)/(17)*-(7)/(25))-((8)/(17)*-(24)/(25)) \n\n\sin(\alpha-\beta)=(105)/(425)-(-(192)/(425)) \n\n \boxed{\sin(\alpha-\beta)=(297)/(425)}$

a miner found three pieces of gold in an under ground mine in 1970 (in 1970 one ounce of gold cost 36.02$). The pieces weighed .080 ounce, .122 ounce, and .096 ounce. Find the total value of gold when it was mined.

Answers

The total value of the mined gold is $ 10.73. This is computed by summing up all the weights of 3 mined gold, 0.80 + 0.122 + 0.096, which will give us 0.298 ounce. Then this will be then multiply be 36.02, the cost of gold, which will give us $10.73.

The total value of gold when it was mined was $10.73.

In 1970, one ounce of gold cost $36.02. A miner found three pieces of gold in an underground mine in 1970.

The pieces weighed 0.080-ounce, 0.122 ounce, and 0.096 ounce. To find the total value of gold when it was mined, you need to multiply the weight of each piece of gold by the price per ounce in 1970 and add the results. To do so, let's find the value of each of the three pieces of gold: Value of first piece = 0.080 ounce × $36.02/ounce = $2.88Value of second piece = 0.122 ounce × $36.02/ounce = $4.39Value of third piece = 0.096 ounce × $36.02/ounce = $3.46Now, add up the value of all three pieces to find the total value of gold: $2.88 + $4.39 + $3.46 = $10.73

Therefore, the total value of gold when it was mined was $10.73.

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According to the Rational Roots Theorem, which statement about f(x) = 25x^7 – x^6 – 5x^4 + x – 49 is true?Any rational root of f(x) is a multiple of –49 divided by a multiple of 25.

Any rational root of f(x) is a multiple of 25 divided by a multiple of –49.

Any rational root of f(x) is a factor of –49 divided by a factor of 25.

Any rational root of f(x) is a factor of 25 divided by a factor of –49.

Answers

Answer:

Any rational root of f(x) is a factor of -49 divided by a factor of 25.

Step-by-step explanation:

The Rational Root Theorems states that :

If the polynomial $P(x)= a _nx^n +{a_(n-1)x}^(n-1)+............{a_(2)x}^(2)+{a_(1)x}^(1)+a_0$ has any rational roots, then they must be in the form of

$\pm (factors of a_0)/(factors of a_n)$

Consider the polynomial

$f(x)=25x^7-x^6-5x^4+x-49$

in this case, we have $a_0=-49$ and $a_n=25$

Any Rational root of f(x) is a factor of $a_0=-49$ divided by a factor of $a_n=25$

According to the Rational Roots Theorem, the statement about f(x) = 25x^7 – x^6 – 5x^4 + x – 49 which is true is:

Any rational root of f(x) is a factor of –49 divided by a factor of 25.

The key points are "factors", and "ratio between the constant term and the coefficient of the highest order (exponent) term"

Two friends leave school at the same time, Sarah is heading due north and Beth is heading due east. One hour later they are 10 miles apart. If Sarah had traveled 8 miles from the school, how many miles had Beth traveled?6 miles
10 miles
8 miles
4 miles

Answers

If you picture this scene in your head, you will get a simple 3x4x5 right triangle. 4 times 2 is 8, 5 times 2 is 10, so Sarah should, by now, have traveled 3 times 2 miles.
Final answer: Sarah has traveled 6 miles after an hour.
Hope this helped! :)

Helping tip: The reason it is not 4 is because they are not traveling west and east away from each other. They are traveling north and east, so they are moving in two directions to form a right triangle, and the hypotenuse would be 10 (miles). The side, or distance, that Beth has traveled is 8. So as this is a typical variation of a 3x4x5 triangle, Sarah will have traveled 6 miles.

wouldn't it actually be 2 miles but since its not listed I would say 4 because if there 10 miles apart and Sarah is 8miles from the school were they both started then 10-8=2. or you can take the two and divide it by 8 and get 4

What is sas congruence?

Answers

The SAS congruence is basically:
S- side
A- angle
S - side

If two triangles have two congruent sides and share one angle (or have one angle in between), then it (the triangles) are considered congruent.

s-side A-angle s-side

The vertices of a quadrilateral ABCD are A(1,1), B(1,5), C(5,5), and D(7,1).Part a) A trapezoid is a four-sided figure with exactly one pair of parallel sides. Is ABCD a trapezoid? Explain your answer.

Part b) You want to transform ABCD into a parallelogram by only moving point B. A parallelogram is a four-sided figure with both pairs of opposite sides parallel. What should be the new coordinates of point B? Explain your answer.

Answers

A. Yes ABCD is a trapezoid ONE way i remember this is trapezoids aerialist always the shape of some square/rectangle w/a triangle

B. You can always do tis on graph paper. So we will be moving point B from (1,5) to............... (-1,5)

Hope this helps!!!

As it stands now, A and D are on the same horizontal line, and B and C are on the same horizontal line. However, we want to move only point B.So, if we keep BC parallel to AD, then we want to have BC the same length as Ad, which is 6. Right now, the length of BC is 5-1= 4. So, to increase the length to 6, staying on the line y=5, we move B to (-1, 5).

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