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3 years ago

Ryan opened a savings account with $1700 that pays no interest. He deposits an

Mathematics

1 answer:

riadik2000 [5.3K]3 years ago

5 0

$1700 + $190(t)
$1700 + $190 (32) = $7780

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Which expression is equivalent to<br> ^3 sqrt 1/1000 c^9 d^12

tamaranim1 [39]

Answer:

$\sqrt[3]{ \frac{1}{1000} {c}^{9} {d}^{12} } \\ = \frac{1}{10} {c}^{3} {d}^{4}$

5 0

3 years ago

Sandra used partial products to find the product of 438 × 17 by multiplying 438 by 1 and 438 by 7 to get 3,066. Find the product

boyakko [2]

Let us determine the product of 438 and 17 by partial products.

Consider $438 = 400+30+8$

and $17 = 10+7$

So, $438 \times 17 = (400+30+8) \times (10+7)$

$438 \times 17 = (400\times 10)+(30 \times 10)+(8 \times 10) +(400 \times 7)+(30 \times 7)+(8 \times 7)$

= $438 \times 17 = (4000+300+80+2800+210+56)$

= 7446

Therefore, the product of 438 and 17 is 7446.

No, Sandra's answer is not correct.

Because she should have expressed 17 as (10+7), then if she multiplied (438 by 10) and (438 by 7). And, then added the results.Then her answer would be correct.

6 0

3 years ago

Evaluate Y=3x-5 when x=6

Harman [31]

7 + (-3x^2) for x = 0 7 + [-3(0)^2]; = 7 + -3(0); = 7 - 0; = 7

6 0

3 years ago

Find the solution(s) to the system of equations represented in the graph.

jolli1 [7]

Answer:

(-3,0) and (0,3)

Step-by-step explanation:

The solution to the system of equation is where the two graphs intersect

The intersect at two points

(-3,0) and (0,3)

3 0

3 years ago

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$

8 0

3 years ago