How do I do this problem?? I don't get it.

notsponge [240]

Answer:

1a - no

1b - yes

1c - no

2 - 1.5 grams of protein

3 - a solution to this equation tells us how many grams of protein/fat there could be depending on how many grams of the other there are. a solution to this is 6 grams of protein and 4 grams of fat.

Step-by-step explanation:

1a

4(5)+9(2)=60

20+18=60

38=60

the equation is false

1b

4(10.5)+9(2)=60

42+18=60

60=60

the equation is true

1c

4(8)+9(4)=60

32+36=60

68=60

the equation is not true

2

plug in 6 for the f value

4p+9(6)=60

4p+54=60

subtract 54 from both sides

4p=6

divide both sides by 4 to get p alone

p=1.5

3

a possible solution can be seen by graphing the equation using the intercepts. the intercepts for this equation are (15,0) and (0,6.6). attached is an image of the graph. the points where the line crosses are possible solutions. the line crosses the point (6,4) on the graph, which represents 6 grams of protein and 4 grams of fat. you can also check this by plugging these values into the equation.

4 0

3 years ago

What is the inverse of (x^2 4)

bazaltina [42]

The inverse function of the given function y = x² + 4 will be y = √(x - 4).

<h3>What is the inverse function?</h3>

Let the function f is given as

y = m(x + a) + c

Then the inverse function of the function f will be given by swapping x with y and y with x.

The definition of simplicity is making something simpler to achieve or grasp while also making it a little less difficult.

The function is given below.

y = x² + 4

The inverse function of the function y = x² + 4 is given as,

Replace x with y and y with x, then we have

x = y² + 4

Solve the equation for y, then the equation is rewritten as,

x = y² + 4

y² = x - 4

y = √(x - 4)

The inverse function of the given function y = x² + 4 will be y = √(x - 4).

More about the inverse function is given below.

brainly.com/question/2541698

#SPJ12

5 0

1 year ago

Plzzz Solve these 7 Questions ASAP !!<br> PLZZZZZZZZZZZ IT'S A REQUEST !!!

e-lub [12.9K]

Answer:

B is an answer.

6 0

3 years ago

13 ft :10 ft 18 ft Find the area of the trapezoid. [? ] square feet Hint: The formula for the area of a trapezoid is: (b17b2). h

Liono4ka [1.6K]

The area is 29473
i just need points

3 0

3 years ago

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