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3 years ago

15

if you needed only 1c of milk, what is your best choice at the grocery store- a quart container, a pint container, or a 1/2 gal

containers?

2 answers:

Artist 52 [7]3 years ago

3 0

A pint container because there are 2 cups in a pint, whereas there are 6 cups in 1/2 gallon and 4 cups in a quart.

Ahat [919]3 years ago

3 0

1 cup = 8 ounces not a lot of milk the smallest container would be your best choice, the pint is the smallest 1 pint = 16 ounces hope this helps

You might be interested in

Graph the linear equation find three points that solve the equation then plot on the graph 2y=-x+2

balu736 [363]

I guess this is the line
$\binom{0}{2} \binom{2}{3} \binom{4}{4}$

7 0

3 years ago

If EF=9x+14, FG=56, and EG=250, find the value of x.

nexus9112 [7]

Answer:

Value of x =20

Step-by-step explanation:

Given: EF = 9x+14 units , FG = 56 units and EG = 250 units.

Segment Addition Postulates states the following for 3 points that are collinear.

i.e, Let three points A, B and C are collinear and B is between A and C.

i,e AC = AB + BC

By Segment addition postulates; solve for x;

EG = EF + FG

Substitute the given values we get;

250 = 9x + 14 + 56

or

250 = 9x + 70

Subtract 70 on both sides we get;

250 -70 = 9x + 70 -70

Simplify:

180 = 9x

Divide both sides by 9 we get;

$\frac{180}{9} = \frac{9x}{9}$

Simplify:

x = 20

Therefore, the value of x =20

7 0

3 years ago

Read 2 more answers

a set v is given, together with definitions of addition and scalar multiplication. determine which properties of a vector space

agasfer [191]

The properties of a vector space are satisfied Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes aren't legitimate are if $v = x ^ 2 1× v=1^ ×x ^ 2 = 1 #V$

Property three does now no longer follow: Suppose that Property three is legitimate, shall we name $v = a * x ^ 2 +bx +c$ the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently $0 = O + v = (O x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a$

= 0 If O is the neuter, then it ought to restore x², but $0+ x² = (0x²+0x+zero) + (x²+0x+zero) = 1.$ This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant

have additive inverse

$Let r= v ×2x ^ 2 + v × 1x +v0 , w= w ×2x ^ 2 + w × 1x +w0 . We have that\\v+w= (vO + wO) ^ x^ 2 +(vl^ × wl)^ x+ ( v 2^ × w2)• w+v= (wO + vO) ^x^ 2 +(wl^ × vl)x+ ( w 2^ ×v2)$

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could use z = 1 then $v = x ^ 2 w = x ^ 2 + 1\\(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1$

v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3

Since 3x ^ 2 +1 ne x^ 2 +3. then the associativity rule doesnt hold.

$(1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 3\\1^ × (x^ 2 +x)+2^ × (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )\\(1^ ×2)^ ×(x^ 2 +x)=2^ × (x ^ 2 + x) = 2x + 2\\1^ × (2^ × (x ^ 2 + x) )=1^ × (2x+2)=2x^ 2 +2x( ne2x+2)$

Property f doesnt observe because of the switch of variables. for instance, if v = x ^ 2 1 × v=1^ × x ^ 2 = 1 #V

Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes arent legitimate.

Step-with the aid of using-step explanation:

Note that each sum and scalar multiplication entails in replacing the order from that most important coefficient with the impartial time period earlier than doing the same old sum/scalar multiplication.

Property three does now no longer follow: Suppose that Property three is legitimate, shall we name v = a × x ^ 2 +bx +c the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently $0 = O + v = (O × x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a$

= zero If O is the neuter, then it ought to restore x², but zero $+ x² = (0x²+0x+zero) + (x²+0x+zero) = 1.$ This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant have additive inverse

Let $r= v × 2x ^ 2 + v × 1x +v0 , w= w2x ^ 2 + w × 1x +w0 . \\We have thatv+w= (vO + wO) ^ x^ 2 +(vl^ wl)^x+ ( v 2^ w2)w+v= (wO + vO) ^ x^ 2 +(wl^ vl)x+ ( w 2^v2)$

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could use $z = 1 then v = x ^ 2 w = x ^ 2 + 1(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3\\Since 3x ^ 2 +1 ne x^ 2 +3.$ then the associativity rule doesnt hold.

Note that each expressions are same because of the distributive rule of actual numbers. Also, you could be aware that his assets holds due to the fact in each instances we 'switch variables twice.

$· (1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 31^ * (x^ 2 +x)+2^ * (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )(1^ * 2)^ * (x^ 2 +x)=2^ * (x ^ 2 + x) = 2x + 21^ * (2^ * (x ^ 2 + x) )=1^ ×* (2x+2)=2x^ 2 +2x( ne2x+2)$

Read more about polynomials :

brainly.com/question/2833285

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8 0

11 months ago

1. Find the digit that makes 3,71_ divisible by 9.

KiRa [710]

<u>QUESTION 1</u>

We want to find the digit that should fill the blank space to make

$3,71-$

divisible by 9.

If a number is divisible by 9 then the sum of the digits should be a multiple of 9.

The sum of the given digits is,

$3 + 7 + 1 = 11$

Since

$11 + 7 = 18$

which is a multiple of 9.

This means that

$3,717$

is divisible by 9.

The correct answer is B

<u>QUESTION 2</u>

The factors of the number 30 are all the numbers that divides 30 exactly without a remainder.

These numbers are ;

$1,2,3,5,6,10,15,30$

The correct answer is A.

<u>QUESTION 3.</u>

We want to find the prime factorization of the number 168.

The prime numbers that are factors of 168 are

$2,3 \: and \: 7$

We can write 168 as the product of these three prime numbers to obtain,

$168={2}^{3}\times 3\times7$

We can also use the factor tree as shown in the attachment to write the prime factorization of 168 as

$168 ={2}^{3}\times 3\times7$

The correct answer is B.

QUESTION 4.

We want to find the greatest common factor of

$140\:\:and\:\:180$

We need to express each of these numbers as a product of prime factors.

The prime factorization of 140 is

$140={2}^{2}\times 5\times7.$

The prime factorization of 180 is

$180={2}^{2} \times{3}^{2}\times5.$

The greatest common factor is the product of the least degree of each common factor.

$GCF={2}^{2}\times5$

$GCF=20$

The correct answer is A.

QUESTION 5.

We want to find the greatest common factor of

$15,30\: and\:60.$

We need to first find the prime factorization of each number.

The prime factorization of 15 is

$15=3\times5.$

The prime factorization of 30 is

$30=2\times 3\times 5.$

The prime factorization of 60 is

$60={2}^{2}\times3 \times5$

The greatest common factor of these three numbers is the product of the factors with the least degree that is common to them.

$GCF=3 \times5$

$GCF=15$

The correct answer is C.

QUESTION 6

We want to determine which of the given fractions is equivalent to

$\frac{3}{8}.$

We must therefore simplify each option,

$A.\: \: \frac{15}{32}=\frac{15}{32}$

$B.\:\:\frac{12}{32}=\frac{4\times 3}{4\times8}=\frac{3}{8}$

$C.\:\:\:\:\frac{12}{24}=\frac{12\times1}{12\times 2}=\frac{1}{2}$

$D.\:\:\frac{9}{32}=\frac{9}{32}$

The simplification shows that

$\frac{12}{32}\equiv \frac{3}{8}$

The correct answer is B.

QUESTION 7.

We want to express

$\frac{10}{22}$

in the simplest form.

We just have to cancel out common factors as follows.

$\frac{10}{22}=\frac{2\times5}{2 \times11}$

This simplifies to,

$\frac{10}{22}=\frac{5}{11}$

The correct answer is C.

QUESTION 8.

We were given that Justin visited $25$ of the $50$ states.

The question requires that we express $25$ as a fraction of $50.$

This will give us

$\frac{25}{50}=\frac{25\times1}{25\times2}$

We must cancel out the common factors to have our fraction in the simplest form.

$\frac{25}{50}=\frac{1}{2}$

The correct answer is C.

QUESTION 9.

We want to write

$2\frac{5}{8}$

as an improper fraction.

We need to multiply the 2 by the denominator which is 8 and add the product to 5 and then express the result over 8.

This gives us,

$2 \frac{5}{8}=\frac{2\times8+5}{8}$

this implies that,

$2\frac{5}{8}=\frac{16+5}{8}$

$2\frac{5}{8}=\frac{21}{8}$

Sarah needed

$\frac{21}{8}\:\:yards$

The correct answer is D.

QUESTION 10

See attachment

QUESTION 11

We wan to write

$3\: and\:\:\frac{7}{8}$

as an improper fraction.

This implies that,

$3+\frac{7}{8}=3\frac{7}{8}$

To write this as a mixed number, we have,

$3\frac{7}{8}=\frac{3\times8+7}{8}$

This implies that,

$3\frac{7}{8}=\frac{24+7}{8}$

This gives

$3\frac{7}{8}=\frac{31}{8}$

The correct answer is B.

QUESTION 12

We want to find the LCM of $30$ and $46$ using prime factorization.

The prime factorization of 30 is $30=2\times 3\times 5$

The prime factorization of 46 is $40=2\times 23$ .

The LCM is the product of the common factors with the highest degrees. This gives us,

$LCM=2\times \times3 5\times 23$

$LCM=690$

The correct answer is D.

QUESTION 13

We want to find the least common multiple of 3,6 and 7.

The prime factorization of $3$ is $3$ .

The prime factorization of 6 is $6=2\times 3$ .

The prime factorization of 7 is $7$ .

The LCM is the product of the common factors with the highest degrees. This gives us,

$LCM=2\times3 \times7$

$LCM=42$ .

The LCM is 42, therefore 42 days will pass before all three bikes will at the park on the same day again.

The correct answer is B.

See attachment for continuation.

6 0

3 years ago

Read 2 more answers

Please help me no need to explain just drop the answers

Bas_tet [7]

Answer:

a) x: 0,1,2,3,4,5,6,7,8

y: 0,8,16,24,32,48,56,64

b) label the horizontal line number of yards. The vertical line is costs, label each unit in 8s.

c) constant of porportionality is 8

d) y=8x

4 0

3 years ago