5.1(3+2.2x)>-14.25-6(1.7x+4)

Answer: No, it is not enough to make 2 batches of Mac and cheese and 3 pizzas.

Step-by-step explanation:

Since we have given that

She needed 6 ounces of cheese and 2 tablespoons of basil for each batch of her Mac and cheese.

Now,

For every two pizzas she needed 16 ounces of cheese and 5 tablespoons of basil.

So, for 2 batches of Mac and cheese she needed = 12 ounces of cheese

and if for 2 pizzas she needed = 16 ounces of cheese

For 1 pizza she needed = 8 ounces of cheese

For 3 pizza she needed = 8× 3=24 ounces of cheese

Total requirement of cheese = 24+ 12=36 ounces

But she only buys 32 ounces package of cheese .

So, it is not enough to make 2 batches of Mac and cheese and 3 pizzas and 4 ounces is still needed i.e.

36-32 =4 ounces .

Lena bought a 32 oz package of cheese, and since 60 ounces is greater than 32 ounces, she does not have enough cheese to make 2 batches of Mac and cheese and 3 pizzas with her current supply.

If Lena has enough cheese to make 2 batches of Mac and cheese and 3 pizzas.

Each batch of Mac and cheese requires 6 ounces of cheese, and she wants to make 2 batches,

so that's 6 ounces * 2

= 12

Each pizza requires 16 ounces of cheese, and she wants to make 3 pizzas,

so that's 16 ounces * 3

= 48

In total, for the Mac and cheese and the pizzas,

Lena needs 12 ounces + 48 ounces

= 60 ounces of cheese.

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2x * x : 2 = 400

x = 20 (side x)

20 * 2 = 80 (side 2x)

√80²+20²=

82.46 (side H)

The length of the sides of the triangle are x = 20 units , 2x = 40 units , and H = 44.72 units.

Given data:

Let the triangle be represented as ΔABC

Now, the measure of the sides of the triangle are:

The measure of AB = 2x

The measure of BC = x

And, the measure of hypotenuse H = H

The area of the triangle is 400 units²

So, 400 = ( 1/2 )(x ) (2x )

x² = 400

x = 20 units

So, the measure of AB = 40 units

And, the hypotenuse is H = √ [ ( 2x )² + ( x )² ]

H = √ ( 1600 + 400 )

H = √2000

H = 44.72 units

Hence, the length of the triangle is solved.

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9x^2 + 27x + 18 . hope i helped

The factors of the quadratic equation are (-3 - √11i) and (-3 + √11i).

A quadratic equation is an algebraic equation of degree 2.

Given a quadratic equation 9x² + 27x + 45 = 0.

We know simplifying a quadratic equation doesn't change the roots of the equation.

∴ We can divide each of the terms by 9 which makes the quadratic equation x² + 3x + 5 = 0.

Now two numbers which multiplied to 5 and add to 3 are not possible for  real numbers, therefore, the roots must be imaginary.

We know Sridhar Acharya's formula which is [tex]\frac{ - b \pm\sqrt{b^2 - 4ac}}{2a}[/tex].

[tex]= \frac{ - 3 \pm\sqrt{3^2 - 4.1.5}}{2.1}[/tex].

[tex]= \frac{ - 3 \pm\sqrt{9 - 20}}{2}[/tex].

[tex]= \frac{ - 3 \pm\sqrt{-11}}{2}[/tex].

[tex]= \frac{ - 3 \pm\sqrt{11}i}{2}[/tex].

Therefore the factors of the quadratic equation are

(-3 - √11i) and (-3 + √11i).

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Here we're moving the decimal point in 0.176 one place to the left.  Thus, x = -1.  Note that 1.76*10^(-1) = 0.176.

Hey there!!

Given :

[tex]0.176 = 1.76 * 10^{x}[/tex]

If we place -1 in place of ' x ', we get the required answer.

Hence, x = -1.

Hope my answer helps!!

Answer:

Step-by-step explanation:

Given that the plane passes through (1,3,3)

Also the plane contains the line x(t)=3t,y(t)=2t,z(t)=4+3t.

This means all points lying in this line will also lie in the plane.

Find out two points on this line.

First point: Let t =0.  Point is (0,0,4)

Next point : Let t = 1: Point is (3,2,7)

Now we have 3 non collinear points (0,0,4) (3,2,7) and (1,3,3) lying on the plane.

Equation of the plane is

[tex]\left[\begin{array}{ccc}x-0&y-0&z-4\\3&2&3\\1&3&-1\end{array}\right] =0[/tex]

Simplify to get

x(-2-9)-y(-3-3)+(z-4)(9-2)=0

i.e -11x+6y+7z-28 =0

11x-6y-7z+28 =0 is the equation of the plane.

The answer is C because AO and BO are equal. And with AO being three times the length of CO, BO should be the length of DO

Answer: Fourth option. 55 mistakes

Solution:

Please, see the attached graph.

Thanks.

The amount of money remaining in Luisa's account is her original balance minus the amount of the check she wrote:

$541.82 - 27.19 = $514.63

The slope we can get from using this y1-y2/x1-x2

so

10-3/-5-1=7/-6=-1 1/6

There's your slope

Apparently, the calculator at the link in your lesson is fully capable of giving you the necessary numbers. My own TI-84 work-alike gives me the account balances, but the rest of the numbers need to be figured.

In 30 years, there are 12×30 = 360 months, or 4×30 = 120 quarters. See the calculator results below. Your table can be filled in using the given information to find the contribution amount. The calculator gives the final balance. The interest amount is found by subtracting the contribution amount from the final balance.

[tex]\begin{array}{cccc}\text{Opt}&\text{Contributions}&\text{Int}&\text{Final Bal}\\1&360\cdot 25=9000&6250.50&15250.50\\2&120\cdot 75=9000&8467.04&17467.04\\3&1000&5489.17&6489.17\end{array}[/tex]

The answer is c since it has a slope of 1/5 and a y-intercept of 1

Remark

So far what you have is the y intercept which according the graph, is (0,1)

The general equation is y = mx + b

If you put the following point (0,1) into the general line formula, you will get b

Givens

y = 1

x = 0

Solve

Substitute the givens into the general line

1 = m*0 + b            The m is no longer there.

b = 1                       The y intercept is 1

So far what you have is

y = mx + 1

The problem tells you that the slope (m) = 1/5

The equation is y = (1/5)x + 1

None of the answers are correct. There is no x anywhere that is right. If x was present, the answer would be C.

Answer:

There are three kind of beta-thalassemia.

Total possible outcome=  3 types of beta- thalassemia + 1 [child will have no trace of disease] = 4

As we know Probability(outcome) = [tex]\frac{\text{ Total favorable outcome}}\text{Total possible outcome}[/tex]

Probability that the child will have beta-thalassemia major =\frac{_{1}^{3}\textrm{C}}{_{1}^{4}\textrm{C}}

     =[tex]\frac{3}{4}[/tex]

   ∴ favorable=[tex]_{1}^{3}\textrm{C}[/tex]

 And , Total =[tex]_{1}^{4}\textrm{C}[/tex]

Probability that the child will have no trace of the disease=[tex]\frac{1}{4}[/tex]

The probability of Mrs. R's child having beta-thalassemia major or no trace of the disease depends on the genetic status of both parents. A 25% chance exists for beta-thalassemia major if both parents are carriers, with a 25% chance for no trace of the disease. Specific probabilities require knowledge of both parents' genetic status.

Beta-thalassemia is a genetic blood disorder that results from a mutation in the beta-globin gene, leading to a deficiency in the synthesis of the beta-chain of hemoglobin. To determine the probability that Mrs. R's child will have beta-thalassemia major or no trace of the disease, we would need to know the genetic status of both parents. If both parents are carriers (heterozygous for the beta-thalassemia trait), there is a 25% chance that the child will inherit two defective genes resulting in beta-thalassemia major, a 50% chance the child will be a carrier like the parents, and a 25% chance the child will inherit two normal genes and thus have no trace of the disease.

If one parent has beta-thalassemia major and the other is a carrier, every child will have at least one defective gene; hence, they all will be at least carriers. If one parent has beta-thalassemia major and the other has normal genes, each child has a 100% chance of being a carrier. Without knowing the specific genetic status of Mrs. R and her partner, we cannot provide exact probabilities.

The appropriate choice is ...

.... A. The student failed to keep both sides of the equation balanced

_____

The one rule I learned in my study of Algebra that makes it all make sense is "keep the equal sign sacred." That is, whatever you do to one side of the equation must also be done to the other side of the equation. (Various teachers show various shortcuts that make it look like you're doing different things to the different sides of the equation. Don't be confused. If the operation can't be put in terms of doing the same thing to both sides, it doesn't work.)

Of course, you can always add 0 or multiply by 1 whenever and wherever you want.

_____

For the above equation, the appropriate solution is

... a + 3 = 9

... a + 3 - 3 = 9 - 3 . . . . . . addition property of equality

... a + 0 = 6 . . . . . . . . . . . .rules of arithmetic

... a = 6 . . . . . . . . . . . . . . . additive identity property of 0

the perimeter is the sum of the 3 sides of the triangle, hence

x + 1 + x + 2 + x + 2 = 32

3x + 5 = 32 ( subtract 5 from both sides )

3x = 27 ( divide both sides by 3 )

x = 9

EF = x + 1 = 9 + 1 = 10

FG = x + 2 = 9 + 2 = 11

EG = x + 2 = 9 + 2 = 11

check : 10 + 11 + 11 = 32 ← perimeter

1. You can make use of the Pythagorean theorem considering "a" to be the longer leg of the largest right triangle. Then the value of "a" can be calculated from

(16+9)² = 15² +a²

625 -225 = a²

√400 = a = 20

2. You can find the altitude of the largest triangle (the length of the unmarked vertical line), then find "a" as the hypotenuse of the medium-sized right triangle.

That altitude is ...

√(15² -9²) = √144 = 12

so the length "a" is ...

a² = 12² +16² = 144 +256 = 400

a = √400 = 20

3. Based on the 15 and 9 dimensions of the smallest right triangle, you can realize that these triangles are multiples of the 3-4-5 right triangle. Then you can use relevant ratios of the side lengths of any of the triangles of which "a" is a part.

a = 4/3×15 = 4/5×(9+16) = 5/4×16 = 20

9/2 would be on the right -7/2 will be on the left hope i helped

Answer:

The easiest way to plot the fraction in the number line is to change the fraction into the decimal then plot it, as decimal is easily detected in the number line as compare to the fraction.

The Fraction [tex]\frac{9}{2} = 4.5[/tex]

and value 4.5 lies exact between 4 and 5.

The Fraction [tex]\frac{-7}{2} = -3.5[/tex]

and value 4.5 lies exact between -4 and -3.

It is shown below.

(f+g)(x) = f(x) + g(x)

Now all you have to do is plug in the equations:

(x/2-3) + (4x^2 + x -4)

Combine like terms:

3x/2-7+4x^2

This is equal to the third equation, so the answer is:

C

To find the sum of the functions f(x) and g(x), combine them like so: (f + g)(x) = (x/2 - 3) + (4x^2 + x - 4), which simplifies to (f + g)(x) = 4x^2 + 3/2x - 7.

Given the functions f(x) = x/2 - 3 and g(x) = 4x^2 + x - 4, the operation (f + g)(x) signifies the summation of the two functions. You simply add the two functions together.

To find (f + g)(x), we combine f(x) and g(x) like so: (f + g)(x) = f(x) + g(x) = (x/2 - 3) + (4x^2 + x - 4).

By combining like terms, we simplify to: (f + g)(x) = 4x^2 + 3/2x - 7.

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Perimeter of the triangle ABC is 18.7 units.

Coordinate of the point A is (4, -1), B is (-1, 4) and C is (0, -3).

Length of the side AB = [tex]\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/tex]

AB = [tex]\sqrt{(-1-4)^{2}+(4+1)^{2}} =\sqrt{50}[/tex]

AB = 7.07 units = 7.1 units ( rounded to the nearest 10th)

Length of the side BC = [tex]\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/tex]

BC = [tex]\sqrt{(0+1)^{2}+(-3-4)^{2}}[/tex]

BC = [tex]\sqrt{50}[/tex]

BC = 7.07 units = 7.1 units ( rounded to the nearest 10th)

Length of the side CA =[tex]\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/tex]

CA = [tex]\sqrt{(4-0)^{2}+(-1+3)^{2}}[/tex]

CA = [tex]\sqrt{20}[/tex]

CA = 4.47 units = 4.5 ( rounded to the nearest 10th)

Perimeter of the triangle ABC = sum of all three sides = (7.1 + 7.1 + 4.5) units = 18.7 units  

AB:BC = 3:2

so AB=3/2BC

AB+BC=AC=20

substitute 3/2BC+BC=20

5/2BC=20

BC=2*20/5=8

AB=3/2*8=12

Apply a:b=c:d as a/b=c/d

AB:BC = 3:2 is AB/BC=3/2

AB=3/2*BC

AC=20=AB+BC

So 20=3/2*BC+BC

1 3/2BC=20

BC=8cm

AB=20-8=12cm

Answer:

Y = 5

Step-by-step explanation:

Since we know six is the value of X we can go ahead and multiply 4 and 6

4 X 6 = 24

Now there are different ways to find the value of Y

How I find the value is I subtracted 24 from 9

That gives you 15

Next you find what number you need to multiply by 3 to get 15

In this case your answer is 5

Then to be extra sure subtract 24 and 15 and you'll get 9

Hope this helps!

y = 5

substitute x = 6 into the equation and solve for y

(4 × 6 ) - 3y = 9

24 - 3y = 9 ( subtract 24 from both sides )

- 3y = - 15 ( divide both sides by - 3 )

y = 5

Answer:

bc

Step-by-step explanation:

The multiplication property of equality tells you that you can multiply both sides of the equation by the same amount. Choose that amount to be bd, the product of the denominators.

[tex]\dfrac{a}{b}=\dfrac{c}{d}\\\\\dfrac{a\cdot bd}{b}=\dfrac{c\cdot bd}{d}\\\\ad\dfrac{b}{b}=bc\dfrac{d}{d}\\\\ad=bc[/tex]

Answer:

The correct options are a and b.

Step-by-step explanation:

It is given that triangle ABC with segment AD drawn from vertex A and intersecting side BC.

Two triangle are called similar triangle if their corresponding sides are proportional or the corresponding interior angle are same.

To prove ΔABC and ΔDBA are similar, we have to prove that corresponding interior angles of both triangle as same.

If segment AD is an altitude of ΔABC, then angle ADB is a right angle.

[tex]\angle BDA=90^{\circ}[/tex]

The opposite angle of hypotenuse is right angle. If segment CB is a hypotenuse, then angle ABC is a right angle.

[tex]\angle BAC=90^{\circ}[/tex]

In triangle ΔABC and ΔDBA

[tex]\angle ABC\cong \angle DBA[/tex]              (Reflexive property)

[tex]\angle BAC\cong \angle BDA[/tex]              (Right angles)

By AA rule of similarity ΔABC and ΔDBA are similar.

Therefore correct options are a and b.

The value of csc(19°) evaluated using cofunctions of complementary angles is equal to sec(71°), since 19° and 71° are complementary angles, and the sine of one angle is equal to the cosine of its complement.

To evaluate csc(19°) using cofunctions of complementary angles, we recall that cosecant (csc) is the reciprocal of the sine function and that the secant (sec) function is the reciprocal of cosine. We also use the fact that complementary angles add up to 90 degrees. Thus, since 19° and 71° are complementary (19° + 71° = 90°), we have:

csc(19°) = 1/sin(19°)

Since sin(19°) is equal to cos(71°) (because sine of an angle is equal to the cosine of its complement), we can write this as:

csc(19°) = 1/cos(71°)

And since sec() = 1/cos(), we can conclude:

csc(19°) = sec(71°)

Answer:

No.

Step-by-step explanation:

You have to imagine (or create a physical model so you can see) that the left and right frame pieces are joined to the top frame piece. The angle each "vertical" piece makes with the horizontal is the sum of the angles shown at each corner: 37° + 50° = 87°.

The angles at each of the corners constitute same-side interior angles where the horizontal transversal crosses the "vertical" lines of each of the side pieces. If the side pieces were parallel, those angles would be supplementary (add to 180°). Instead, they add to 87° + 87° = 174° ≠ 180°. Hence the "verticals" are not parallel.

_____

Usually, a "square" frame has right-angle corners. Here, the corner angles add to 87°, a few degrees short of a right angle. The effect will be to make the side pieces closer together at the bottom than they are at the top. (Parallel lines are the same distance apart everywhere.)

d = 46t

           d = 46(7)

           d = 322 miles                          (7, 322)

            d = 46(12)

            d =   552 miles                      (12, 552)

            d = 46(18)

           d = 828 miles                          (18, 828)    

Answer:

12.1

Step-by-step explanation:

We use the formula for the distance between two arbitrary points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] in the xy-coordinate plane, that is:

[tex]d=\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}[/tex]

So, replacing the points [tex]M=(2,16)[/tex] and [tex]M_2=(-3,5)[/tex], we obtain:

[tex]d=\sqrt{(2-(-3))^2 + (16-5)^2}\\d=\sqrt{(2+3)^2 + (16-5)^2}\\d=\sqrt{5^2 + 11^2}\\d=\sqrt{146 }=12.083045... \simeq 12.1[/tex]

that is the answer.

note: observe that we only use the coordinates between the two midpoints and not the point J.

(b)

A polynomial has the form

[tex]a_{0}[/tex][tex]x^{n}[/tex] ± [tex]a_{1}[/tex][tex]x^{n-1}[/tex] + ....

the only expression fitting this description is : x² + 2x + 7