30 TENS MINUS 3 TENS 3 TENTHS

(a) Reflect the point (7, -9) across the y-axis:

UkoKoshka [18]

Answer:

(-7,-9) and (7,9)

Step-by-step explanation:7,-9 so we do across the x so you need a graph for this now x goes left so we plot the coordinates in the third box 7,-9 is in the 4 box so after we plotted that you can see it reflects.Now y coordinate since the y line goes up we plot the coordinates on top of the 4 box where our original coordinates stay now it should be 7,9 the y becomes positive which is 9.

This is not a web answer i answered myself took me some time!

8 0

3 years ago

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Pleaseeeee helpppppp <333

alexandr1967 [171]

Step-by-step explanation:

64+72+4x+20=180(Sum of angles of a triangle )
156+4x=180
4x=180-156
4x=24
X=24/4
X=6

Therefore The value of X is 6 degree.

6 0

2 years ago

only 6 of the 75 trees in the park are at leat 30 feet tall. what percent of the trees are over 30 feet?

ipn [44]

Answer:

8 %

Step-by-step explanation:

To find the percent of trees that are over 30 ft tall, take the number of trees that are over 30 ft tall over the number of tree

Percent = trees over 30 ft tall/ total number of trees

= 6/75

=.08

Change this to a percent by multiply by 100%

.08 * 100%

8 %

4 0

3 years ago

* please help 20 points !! * which statement can be proved if you are given that SK≌LR?

SOVA2 [1]

I believe it would be D) ST ≈ TR

6 0

3 years ago

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Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

3 years ago