All categories

3 years ago

PLEASE HELP! I WILL GIVE BRAINLIEST! Determine whether the statement is always, sometimes, or never true.

Mathematics

1 answer:

hammer [34]3 years ago

5 0

Answer:

13 True

14 Sometimes

15 Never

Step-by-step explanation:

All equilateral triangles must have 60 degree angles

Sometimes a acute triangle can have 2 acute angles

Obtuse angles take up majority of the mesuares in the triangle. All triangles must have 3 angles equal up to 180 degrees

You might be interested in

You can buy 5 cans for green beans at the Village Markets for $3.70. You can buy 10 of the same cans of beans at Sam’s Club for

mihalych1998 [28]

The place with the best buy between village market and Sam's club is Sam's club at $0.59 per can.

Village market:

Green beans = 5 cans
Total cost = $3.70

Unit rate = Total cost / green beans

= 3.70 / 5

= $0.74 per can

Sam's club:

Green beans = 10 cans
Total cost = $5.90

Unit rate = Total cost / green beans

= 5.90/10

= $0.59 per can

Therefore, the place with the best buy between village market and Sam's club is Sam's club at $0.59 per can.

Learn more about unit rate:

brainly.com/question/4895463

#SPJ1

6 0

1 year ago

25. If the scale factor of two similar solids is 3:16, what is the ratio of their corresponding areas and volumes? (1 point) 3:2

vitfil [10]

I'm not sure about #25 but #26 is C. 144

3 0

3 years ago

Read 2 more answers

What is 1+5 please show your work

julsineya [31]

1 plus 5 is six 1+5=6

6 0

3 years ago

Read 2 more answers

Perfect squares. Solve each equation. Check the solutions. g^2+2/3g+1/9=0

Yakvenalex [24]

G^2+2=0,
g^2=-2
g= sqrt-2, not possible.
3g+1=0
3g=-1
g=-1/3
9 cannot be equal to zero for the last one.

6 0

3 years ago

Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.

lakkis [162]

Answer:

$2*sin(x)+y*cos(x)-cos(y)=C_1$

Step-by-step explanation:

Let:

$P(x,y)=2*cos(x)-y*sin(x)$

$Q(x,y)=cos(x)+sin(y)$

This is an exact differential equation because:

$\frac{\partial P(x,y)}{\partial y} =-sin(x)$

$\frac{\partial Q(x,y)}{\partial x}=-sin(x)$

With this in mind let's define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x}=P(x,y)$

and

$\frac{\partial f(x,y)}{\partial y}=Q(x,y)$

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y)

$f(x,y)=\int\ 2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)$

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}$

Now, let's replace the previous result into $\frac{\partial f(x,y)}{\partial y}=Q(x,y)$ :

$cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)$

Solving for $\frac{dg(y)}{dy}$

$\frac{dg(y)}{dy}=sin(y)$

Integrating both sides with respect to y:

$g(y)=\int\ sin(y) \, dy =-cos(y)$

Replacing this result into f(x,y)

$f(x,y)=2*sin(x)+y*cos(x)-cos(y)$

Finally the solution is f(x,y)=C1 :

$2*sin(x)+y*cos(x)-cos(y)=C_1$

7 0

3 years ago