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oksano4ka [1.4K]
3 years ago
13

What is the answer to the math problem

Mathematics
2 answers:
ale4655 [162]3 years ago
3 0
Is this the right answer

sergey [27]3 years ago
3 0

Answer:

-4.97

Step-by-step explanation:

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In triangle RST, RS = 6 cm, ST = 7 cm, and RT = 8 cm. What is the approximate measure of the largest angle in the triangle?
zubka84 [21]
<span>the approximate measure of the largest angle in the triangle is 75.5 degrees</span>


4 0
3 years ago
Read 2 more answers
Convert 7 pounds 3 ounces to kilograms.
Cerrena [4.2K]

Answer:

3.3

Step-by-step explanation:

7lb 3oz

3oz/16oz=.1875lb

7lb+.1875lb=7.1875lb

7.1875lb/2.2kg=3.27kg

8 0
3 years ago
In the year 2002, a company made $3.3 million in profit. For each consecutive year after that, their profit increased by 5%. How
Anton [14]

Answer:

The profit would be 7.7 million in 2007

Step-by-step explanation:

3 0
3 years ago
The sum of three numbers is 95. The first number is 5 more than the third. The second number is 4 times the third. What are the
kotykmax [81]

Answer:

20, 60, 15

Step-by-step explanation:

Let the third number be n, then

The first number is n + 5 and the second number is 4n

Sum the 3 numbers and equate to 95, that is

n + 5 + 4n + n = 95 ← collect like terms on left side

6n + 5 = 95 ( subtract 5 from both sides )

6n = 90 ( divide both sides by 6 )

n = 15

Thus

The first number is 15 + 5 = 20

The second number is 4 × 15 = 60

The third number is 15

4 0
3 years ago
Find the volume of the solid under the plane 5x + 9y − z = 0 and above the region bounded by y = x and y = x4.
svp [43]
<span>For the plane, we have z = 5x + 9y

For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.

Since x > x^4 for y in [0, 1],

The volume of the solid equals

\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx  \\  \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx  \\  \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\  \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0

=\frac{19}{6} - \frac{5}{6} - \frac{1}{2} =\bold{ \frac{11}{6} \ cubic \ units}</span>
8 0
3 years ago
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