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3 years ago

What is the value of expression below when c = 5 and d=4 6c²-5d + 8

Mathematics

2 answers:

valkas [14]3 years ago

6 0

Answer:138

Step-by-step explanation:

Alisiya [41]3 years ago

5 0

Answer:138

Step-by-step explanation:

First plug in the c and d values into the expression:

6(5)^2-5(4)+8

Remember PEMDAS, so you do 5^2 first getting you 25.

6(25)-5(4)+8

Continue with PEMDAS and multiply -5 and 4, and that gets you -20.

150-20+8

Now you simplify the expression to get: 138.

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Simplify. Assume that no denominator is equal to zero. 4^12/4^5

Georgia [21]

Answer:

16,384 or 2^14

Step-by-step explanation:

4^12= 4*4*4*4*4*4*4*4*4*4*4*4=16,777,216

4^5= 4*4*4*4*4=1024

16,777,216/1024=16,384

16,384=2^14

4 0

3 years ago

Eight less than the product of 7 and a number is 3. Use the variable w for the unknown number

Tpy6a [65]

W = unknown number
7w - 8 = 3 (product of 7 and a number, w, subtract 8) is 3
7w = 11
w = 11/7

8 0

3 years ago

Read 2 more answers

For the rectangle provided, the length of this rectangle is 5 mm longer than its width. The perimeter of this rectangle is more

olya-2409 [2.1K]

A) length= w+5
width=w
perimeter=42mm
B) 2(w+5)+2w>42
2w+10+2w>42
4w>32
w>8
The width should be greater than 8 and the length should be more than 13.

Hope this helps!

3 0

3 years ago

Criterion: Identify IV, DV, and hypotheses and evaluate the null hypothesis for an independent samples t test. Data: Use the inf

Papessa [141]

The table is missing in the question. The table is provided here :

Group 1 Group 2

34.86 64.14 mean

21.99 20.46 standard deviation

7 7 n

Solution :

a). The IV or independent variable = Group 1

The DV or the dependent variable = Group 2

b).

$H_0: \mu_1 = \mu_2$

$H_a:\mu_1 < \mu_2$

Therefore, $t = \frac{\bar x_1 - \bar x_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{S_2^2}{n_2}}}$

$t = \frac{34.86 - 64.14}{\sqrt{\frac{21.99^2}{7}+\frac{20.46^2}{7}}}$

t = -2.579143

Now, $df = min(n_1 - 1, n_2 - 1)$

df = 7 - 1

= 6

Therefore the value of p :

$=T.DIST(-2.579143,6,TRUE)$

= 0.020908803

The p value is 0.0209

$p< 0.05$

So we reject the null hypothesis and conclude that $\mu_1 < \mu_2$

7 0

3 years ago

The oblique pyramid has a square base. What is the volume of the pyramid? 2.5 cm3 5 cm3 6 cm3 7.5 cm3

Olenka [21]

Take a look at the attachment below. It fills in for the attachment that wasn't provided in the question -

An oblique pyramid is one that has a top not aligned with the base. Due to this, the height of the pyramid connects with two vertices at its ends to form a right angle present outside the pyramid, knowing that it is always perpendicular to the base. There is no difference between the calculations of the volume of an oblique pyramid and a pyramid however -

$\\Base Area = 2 cm * 2 cm = 4 cm^2,\\Volume ( Pyramid ) = 1 / 3 * ( Base Area ) * ( Height ),\\Volume = 1 / 3 * ( 4 ) * ( 3.75 ),\\-------------------------\\Volume = 5 cm^3$

And thus, you're solution is 5 cm^3, or in other words option b!

3 0

3 years ago