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3 years ago

13

Please answer fast!!!According to the Triangle Sum Theorem, which combinations of side lengths would make a possible triangle? C

hoose all that apply.
Group of answer choices

3, 4, 3.606

3, 6, 9.2

2, 5, 7

3, 4, 5

2 answers:

saw5 [17]3 years ago

7 0

Answer:

Is there a picture of it?

Step-by-step explanation:

worty [1.4K]3 years ago

5 0

Answer:

3, 4, 5 would be the correct answer.

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Can Someone help me

mrs_skeptik [129]

Answer:

1. congruent

2. congruent

3. not congruent

4. not congruent

5. congruent

6. congruent

7. congruent

8. congruent

9. congruent

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Which graph represents a function?<br> X

gayaneshka [121]

If you added more context then I would probably answer your question

5 0

3 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.

sweet-ann [11.9K]

Answer:

2.794

Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then

$\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy$

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

$\bf G(x,y) = (x, y, 3-x^2-y^2)$

with 0≤ x≤1 and 0≤ y≤1

So

$\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)$

we also have

$\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)$

and so

$\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2$

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy$

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

5 0

3 years ago

How many students were surveyed about their mode of transportation to school?

AlekseyPX

Answer:

Total number of students surveyed about their mode of transportation to school is 400.

Step-by-step explanation:

Here, given student who uses car as their mode of transport = 120 students

We have to find the total number of students who are surveyed.

Also, since complete circle represents 100% of data,

Total students who uses at least one mode of transport out of foot, bicycle, bus or car let it be x students.

Total percentage of students that uses bus, bicycle and on foot = (20 +15+35)% = 70%

Thus, percentage of students that uses car = 100 - 70 = 30%

Thus, 30 % of total students = 120

Mathematically, 30% of x = 120

Solving for x, we get,

$\frac{30}{100} \times x=120$

$\Rightarrow x=120 \times \frac{100}{30}$

$\Rightarrow x=400$

Thus, total number of students surveyed about their mode of transportation to school is 400.

8 0

3 years ago

Need help have no idea how to do this

Anna11 [10]

Number one is parallel

4 0

3 years ago