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3 years ago

A recipe calls for 3 cups of sugar for every 5 cups of water. How many cups of sugar are needed for 25 cups of water?

Mathematics

1 answer:

nasty-shy [4]3 years ago

7 0

The answer is 15 cups

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NEED HELP WITH THESE

faltersainse [42]

Step-by-step explanation:

1 .f(x)= $x^{2}$ -x+1

f(-1)= (-1)-(-1)+1

f(-1)= 1+1+1

= 3

3.f(x)= $x^{2}$ -x+1

f( 1)=(1)-1+1

f( 1)=1-1+1

5. f(x)= $x^{2}$ -x+1

f(3)= $(3)^{2}$ -3+1

= 9-3+1

2. g(x) = 5 - 3x

g( -8)=5-3(-8)

g( -8)=5+24

=29

4.g(x) = 5 - 3x

g(5)=5-3(5)

g(5)=5-15

=-10

6.g(x) = 5 - 3x

g(-3)=5-3(-3)

g(-3)=5+9

=14

3 0

3 years ago

PLs help 50 PTS!!!!! PLEASE ILL GIVE BRAINLIEST!!!!!

Nookie1986 [14]

Answer:

$\large\boxed{y=\dfrac{1}{4}x^2-x-4}$

Step-by-step explanation:

The equation of a parabola in vertex form:

$y=a(x-h)^2+k$

(h, k) - vertex

The focus is

$\left(h,\ k+\dfrac{1}{4a}\right)$

We have the vertex (2, -5) and the focus (2, -4).

Calculate the value of a using $k+\dfrac{1}{4a}$

k = -5

$-5+\dfrac{1}{4a}=-4$ add 5 to both sides

$\dfrac{1}{4a}=1$ multiply both sides by 4

$4\!\!\!\!\diagup^1\cdot\dfrac{1}{4\!\!\!\!\diagup_1a}=4$

$\dfrac{1}{a}=4\to a=\dfrac{1}{4}$

Substitute

$a=\dfrac{1}{4},\ h=2,\ k=-5$

to the vertex form of an equation of a parabola:

$y=\dfrac{1}{4}(x-2)^2-5$

The standard form:

$y=ax^2+bx+c$

Convert using

$(a-b)^2=a^2-2ab+b^2$

$y=\dfrac{1}{4}(x^2-2(x)(2)+2^2)-5\\\\y=\dfrac{1}{4}(x^2-4x+4)-5$

use the distributive property: a(b+c)=ab+ac

$y=\left(\dfrac{1}{4}\right)(x^2)+\left(\dfrac{1}{4}\right)(-4x)+\left(\dfrac{1}{4}\right)(4)-5\\\\y=\dfrac{1}{4}x^2-x+1-5\\\\y=\dfrac{1}{4}x^2-x-4$

3 0

4 years ago

Please help me with this math problem Evaluate C/2 - 3 + 6/D when c=14 and d=3

lbvjy [14]

$\frac{c}{2} - 3 + \frac{6}{d } = \\ = \frac{14}{2} - 3 + \frac{6}{3} = \\ = 7 - 3 + 2 = \\ = 6$

7 0

4 years ago

Read 2 more answers

If a storm window has an area of 448 square inches, what are the dimensions if the window is 12 inches higher than it is wide (w

pantera1 [17]

Let width = w then we have the equation:-

w(w + 12) = 448

w^2 + 12w - 448 = 0

(w - 16)(w + 28) = 0

w = 16

Answer: the dimensions of the window are width 16 by height 28 inches

4 0

3 years ago

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.

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