1 question only (find the area)

It takes Andre 4 minutes to swim 5 laps.

soldier1979 [14.2K]

It takes Andre 1.25 minutes to swim a lap.

6 0

2 years ago

Which triangle congruence postulates can be used to prove that Triangle ABD is congruent to Triangle CDB in Rectangle ABCD?

kramer

Answer:

SSS, SAS, ASA, AAS, HL

Step-by-step explanation:

1. SSS (side side side) says if 3 sides of one triangle are congruent to 3 sides of another triangle, then the 2 triangles are congruent.

2. SAS (side angle side) says if 2 sides and the included angle of a triangle are congruent to 2 sides and the included angle of another triangle, then the 2 triangles are congruent.

3. ASA (angle side angle) says if 2 angles and the included side of a triangle are congruent to 2 angles and the included side of another triangle, then the 2 triangles are congruent.

4. AAS (angle angle side) says if 2 angles and the none included side of one triangle are congruent to the corresponding parts of another triangle, then the 2 triangles are congruent.

5 HL (hypotenuse leg) says if 2 right triangles that have a congruent hypotenuse and a corresponding congruent leg, then the 2 triangles are congruent.

Because it is a rectangle, the sides are equal, and they share the same hypotenuse.

4 0

2 years ago

-11=b\10-10 so what does b equal

larisa [96]

So b/10 has to be -1, that means b is -10

5 0

3 years ago

Write an equation that is perpendicular to x-3y=2 and contains the point (2,4)

Kisachek [45]

perp to x-3y=2 thru (2,4)

For perpendicular we swap the coefficients on x and y, negating one

3x + y = some constant

We get the constant in the obvious way from the point

3x + y = 3(2) + 4 = 6 + 4 = 10

Answer: 3x + y = 10

6 0

3 years ago

What is the best approximation of the projection of (5,-1) onto (2,6)?

Hatshy [7]

Answer:

Hence, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\sqrt{10} }{5}$ , and the vector projection of $\vec a$ onto $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$ .

Step-by-step explanation:

We have given two points $(5, -1)$ and $(2, 6)$ .

Let, $\vec a=5\hat {i}-\hat {j}$ and $\vec b= 2\hat {i}+6\hat{j}$ .

and we have calculate the projection of $\vec a$ onto $\vec b$ .

Now,

For the calculation of projection, first we need to calculate the dot product of $\vec a$ and $\vec b$ .

$\vec a.\vec b=(5\hat {i}-\hat{j}).(2\hat{i}+6\hat{j})$

$=10-6$

$=4$

then, we have to calculate the magnitude of $\vec b$ .

$\mid {\vec {b}}\mid$ = $\sqrt{2^{2}+6^{2} }$ = $\sqrt{40}$ = $2\sqrt{10}$ .

Now, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\vec a.\vec b}{\mid b\mid}$

= $\frac{4}{2\sqrt{10} }$ $\frac{2}{\sqrt{10} } \times\frac{\sqrt{10} }{\sqrt{10} } =\frac{2\sqrt{10} }{10} = \frac{\sqrt{10} }{5}$

and the vector projection of $\vec a$ onto $\vec b$ = $\frac{\vec a. \vec b}{\mid\vec b \mid^{2} } . \vec b$

= $\frac{4}{40} . (2\hat i+ 6\hat j)$

= $\frac{1}{5} \hat i+\frac{3}{5} \hat j$

Hence, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\sqrt{10} }{5}$ , and the vector projection of $\vec a$ onto $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$ .

6 0

3 years ago